CodeForces Round 199 Div2

完了,这次做扯了,做的时候有点发烧,居然只做出来一道题,差点被绿.

My submissions

 

 

#	When	Who	Problem	Lang	Verdict	Time	Memory
4434550	Sep 9, 2013 11:57:20 AM	OIer	E - Xenia and Tree	GNU C++	Accepted	842 ms	4260 KB
4434547	Sep 9, 2013 11:56:11 AM	OIer	E - Xenia and Tree	GNU C++	Wrong answer on test 4	280 ms	3400 KB
4434544	Sep 9, 2013 11:54:50 AM	OIer	E - Xenia and Tree	GNU C++	Memory limit exceeded on test 4	966 ms	262100 KB
4434534	Sep 9, 2013 11:50:15 AM	OIer	E - Xenia and Tree	GNU C++	Memory limit exceeded on test 4	748 ms	262100 KB
4434506	Sep 9, 2013 11:33:28 AM	OIer	E - Xenia and Tree	GNU C++	Time limit exceeded on test 19	5000 ms	4200 KB
4434474	Sep 9, 2013 11:15:59 AM	OIer	E - Xenia and Tree	GNU C++	Accepted	748 ms	3600 KB
4434179	Sep 9, 2013 8:45:13 AM	OIer	D - Xenia and Dominoes	GNU C++	Accepted	30 ms	600 KB
4431354	Sep 8, 2013 1:42:22 PM	OIer	C - Cupboard and Balloons	GNU C++	Accepted	30 ms	0 KB
4431256	Sep 8, 2013 1:16:19 PM	OIer	C - Cupboard and Balloons	GNU C++	Wrong answer on test 17	30 ms	0 KB
4431247	Sep 8, 2013 1:14:03 PM	OIer	B - Xenia and Spies	GNU C++	Accepted	124 ms	1200 KB
4431244	Sep 8, 2013 1:13:33 PM	OIer	B - Xenia and Spies	GNU C++	Wrong answer on test 1	0 ms	1300 KB
4431210	Sep 8, 2013 1:00:11 PM	OIer	B - Xenia and Spies	GNU C++	Wrong answer on test 3	30 ms	1100 KB

Judgement protocol

---

Test: # , time:   ms., memory:   KB, exit code:  , checker exit code:  , verdict:  

Input

Output

Answer

Checker Log

← →

 

My contest submissions

 

 

 

#	When	Who	Problem	Lang	Verdict	Time	Memory
4424854	Sep 7, 2013 1:53:25 PM	OIer	B - Xenia and Spies	GNU C++	Wrong answer on pretest 2	30 ms	1100 KB
4424797	Sep 7, 2013 1:52:36 PM	OIer	B - Xenia and Spies	GNU C++	Wrong answer on pretest 1	0 ms	1300 KB
4422952	Sep 7, 2013 1:21:08 PM	OIer	A - Xenia and Divisors	GNU C++	Accepted	30 ms	400 KB
4422863	Sep 7, 2013 1:19:32 PM	OIer	C - Cupboard and Balloons	GNU C++	Hacked	30 ms	0 KB
4422366	Sep 7, 2013 1:11:51 PM	OIer	C - Cupboard and Balloons	GNU C++	Wrong answer on pretest 6	30 ms	0 KB
4420785	Sep 7, 2013 12:50:12 PM	OIer	C - Cupboard and Balloons	GNU C++	Wrong answer on pretest 6	30 ms	0 KB
4417237	Sep 7, 2013 12:10:37 PM	OIer	A - Xenia and Divisors	GNU C++	Hacked	30 ms	400 KB

A. Xenia and Divisors

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Xenia the mathematician has a sequence consisting of n (n is divisible by 3) positive integers, each of them is at most 7. She wants to split the sequence into groups of three so that for each group of three a, b, c the following conditions held:

• a < b < c;
• a divides b, b divides c.

Naturally, Xenia wants each element of the sequence to belong to exactly one group of three. Thus, if the required partition exists, then it has groups of three.

Help Xenia, find the required partition or else say that it doesn't exist.

Input

The first line contains integer n (3 ≤ n ≤ 99999) — the number of elements in the sequence. The next line contains n positive integers, each of them is at most 7.

It is guaranteed that n is divisible by 3.

Output

If the required partition exists, print groups of three. Print each group as values of the elements it contains. You should print values in increasing order. Separate the groups and integers in groups by whitespaces. If there are multiple solutions, you can print any of them.

If there is no solution, print -1.

Sample test(s)

Input

6

1 1 1 2 2 2

Output

-1

Input

6

2 2 1 1 4 6

Output

1 2 4

1 2 6

题意:给出一个由1~7组成的序列,将它们分成三个三个一组,要求每组内的数a,b,c满足a<b<c且a整除b,b整除c.如果能办到,输出一组可行解,否则输出-1.

思路:显然只有{1,2,4},{1,2,6},{1,3,6}这3种可能,只需记录每种数字的个数,按需分配即可.

判定为-1的条件:

1.1的个数不是总数的1/3

2.4的个数大于2的个数

3.3的个数大于6的个数

除此之外还有重要的一点,也是我被Hack的原因:数列中不能出现5和7.

/*1 2 4
1 2 6
1 3 6*/
#include<stdio.h>
#include<string.h>
int a[],w[];
int N;
int main()
{
    //freopen("input.txt","r",stdin);
    while (scanf("%d",&N)!=EOF)
    {
        memset(w,,sizeof(w));
        for (int i=;i<=N;i++)
        {
            scanf("%d",&a[i]);
            w[a[i]]++;
        }
        if (w[]!=(N/) || w[]+w[]!=w[]+w[] || w[]<w[] || w[]<w[] || w[] || w[])
        {
            printf("-1\n");
            continue;
        }
        for (int i=;i<=w[];i++) printf("1 2 4\n");
        for (int i=;i<=w[]-w[];i++) printf("1 2 6\n");
        for (int i=;i<=w[];i++) printf("1 3 6\n");
    }
    return ;
}

B. Xenia and Spies

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Xenia the vigorous detective faced n (n ≥ 2) foreign spies lined up in a row. We'll consider the spies numbered from 1 to n from left to right.

Spy s has an important note. He has to pass the note to spy f. Xenia interrogates the spies in several steps. During one step the spy keeping the important note can pass the note to one of his neighbours in the row. In other words, if this spy's number is x, he can pass the note to another spy, either x - 1 or x + 1 (if x = 1 or x = n, then the spy has only one neighbour). Also during a step the spy can keep a note and not pass it to anyone.

But nothing is that easy. During m steps Xenia watches some spies attentively. Specifically, during step t_i (steps are numbered from 1) Xenia watches spies numbers l_i, l_i + 1, l_i + 2, ..., r_i (1 ≤ l_i ≤ r_i ≤ n). Of course, if during some step a spy is watched, he can't do anything: neither give the note nor take it from some other spy. Otherwise, Xenia reveals the spies' cunning plot. Nevertheless, if the spy at the current step keeps the note, Xenia sees nothing suspicious even if she watches him.

You've got s and f. Also, you have the steps during which Xenia watches spies and which spies she is going to watch during each step. Find the best way the spies should act in order to pass the note from spy s to spy f as quickly as possible (in the minimum number of steps).

Input

The first line contains four integers n, m, s and f (1 ≤ n, m ≤ 10⁵; 1 ≤ s, f ≤ n; s ≠ f; n ≥ 2). Each of the following m lines contains three integers t_i, l_i, r_i (1 ≤ t_i ≤ 10⁹, 1 ≤ l_i ≤ r_i ≤ n). It is guaranteed that t₁ < t₂ < t₃ < ... < t_m.

Output

Print k characters in a line: the i-th character in the line must represent the spies' actions on step i. If on step i the spy with the note must pass the note to the spy with a lesser number, the i-th character should equal "L". If on step i the spy with the note must pass it to the spy with a larger number, the i-th character must equal "R". If the spy must keep the note at the i-th step, the i-th character must equal "X".

As a result of applying the printed sequence of actions spy s must pass the note to spy f. The number of printed characters k must be as small as possible. Xenia must not catch the spies passing the note.

If there are miltiple optimal solutions, you can print any of them. It is guaranteed that the answer exists.

Sample test(s)

Input

3 5 1 3

1 1 2

2 2 3

3 3 3

4 1 1

10 1 3

Output

XXRR

题意:有n个间谍站成一排,编号为S的间谍需要将情报传给编号为F的间谍.审讯者会来巡逻m次,时间分别是Ti,每次查看编号从Li到Ri的间谍.此时这些间谍既不能传出情报也不能接受情报.问最少多长时间将情报传给F.在每一秒持有情报的间谍可以选择不动(X),向右传(R),向左传(L),每次操作需要1单位时间.输出耗时最短的一组可行方案.

思路:纯模拟,关键是理解题意.

#include<stdio.h>
#include<string.h>
int cur,dir,CT;
int T[],L[],R[];
void go()
{
    if (dir==) putchar('R');
    else putchar('L');
    cur+=dir;
}
int main()
{
    int N,M,S,F;
    while (scanf("%d%d%d%d",&N,&M,&S,&F)!=EOF)
    {
        memset(T,,sizeof(T));
        for (int i=;i<=M;i++) scanf("%d%d%d",&T[i],&L[i],&R[i]);
        cur=S,dir=S<F ? :-,CT=;
        for (int i=;i<=0x7FFFFFFF;i++)
        {
            if (cur==F) break;
            if (i==T[CT])
            {
                if ((L[CT]<=cur && cur<=R[CT]) || (L[CT]<=cur+dir && cur+dir<=R[CT])) putchar('X');
                else go();
                CT++;
            }
            else go();
        }
        printf("\n");
    }
    return ;
}

C. Cupboard and Balloons

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A girl named Xenia has a cupboard that looks like an arc from ahead. The arc is made of a semicircle with radius r (the cupboard's top) and two walls of height h (the cupboard's sides). The cupboard's depth is r, that is, it looks like a rectangle with base r and height h + r from the sides. The figure below shows what the cupboard looks like (the front view is on the left, the side view is on the right).

Xenia got lots of balloons for her birthday. The girl hates the mess, so she wants to store the balloons in the cupboard. Luckily, each balloon is a sphere with radius . Help Xenia calculate the maximum number of balloons she can put in her cupboard.

You can say that a balloon is in the cupboard if you can't see any part of the balloon on the left or right view. The balloons in the cupboard can touch each other. It is not allowed to squeeze the balloons or deform them in any way. You can assume that the cupboard's walls are negligibly thin.

Input

The single line contains two integers r, h (1 ≤ r, h ≤ 10⁷).

Output

Print a single integer — the maximum number of balloons Xenia can put in the cupboard.

Sample test(s)

Input

1 1

Output

3

Input

1 2

Output

5

Input

2 1

Output

2

题意:有一个盒子,下部是一个R*2R*H的长方体,上部是一个半径为R高为R的半圆柱,见图.求在不挤压的情况下最多能放进多少个直径为R的球.

思路:首先从底向上两个两个放,直到距离长方体的顶端距离小于R.计算出剩下的高度,当剩余高度小于R/2时,剩下的空间只能放下一个.当剩余高度在[R/2,sqrt(3)R/2)时,剩余空间能放下两个球.当剩余高度大于等于sqrt(3)R/2时,剩余空间能放下三个球.读者可以自行画图计算.

#include<math.h>
#include<stdio.h>
int main()
{
    int R,H;
    while (scanf("%d%d",&R,&H)!=EOF)
    {
        int ans=H/R*;
        H-=ans*R/;
        if (H*<R) ans++;
        else
        {
            if ((double)(*H)<sqrt(3.0)*(double)R) ans+=;
            else ans+=;
        }
        printf("%d\n",ans);
    }
    return ;
}

D. Xenia and Dominoes

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Xenia likes puzzles very much. She is especially fond of the puzzles that consist of domino pieces. Look at the picture that shows one of such puzzles.

A puzzle is a 3 × n table with forbidden cells (black squares) containing dominoes (colored rectangles on the picture). A puzzle is called correct if it meets the following conditions:

• each domino occupies exactly two non-forbidden cells of the table;
• no two dominoes occupy the same table cell;
• exactly one non-forbidden cell of the table is unoccupied by any domino (it is marked by a circle in the picture).

To solve the puzzle, you need multiple steps to transport an empty cell from the starting position to some specified position. A move is transporting a domino to the empty cell, provided that the puzzle stays correct. The horizontal dominoes can be moved only horizontally, and vertical dominoes can be moved only vertically. You can't rotate dominoes. The picture shows a probable move.

Xenia has a 3 × n table with forbidden cells and a cell marked with a circle. Also, Xenia has very many identical dominoes. Now Xenia is wondering, how many distinct correct puzzles she can make if she puts dominoes on the existing table. Also, Xenia wants the circle-marked cell to be empty in the resulting puzzle. The puzzle must contain at least one move.

Help Xenia, count the described number of puzzles. As the described number can be rather large, print the remainder after dividing it by 1000000007 (10⁹ + 7).

Input

The first line contains integer n (3 ≤ n ≤ 10⁴) — the puzzle's size. Each of the following three lines contains n characters — the description of the table. The j-th character of the i-th line equals "X" if the corresponding cell is forbidden; it equals ".", if the corresponding cell is non-forbidden and "O", if the corresponding cell is marked with a circle.

It is guaranteed that exactly one cell in the table is marked with a circle. It is guaranteed that all cells of a given table having at least one common point with the marked cell is non-forbidden.

Output

Print a single number — the answer to the problem modulo 1000000007 (10⁹ + 7).

Sample test(s)

Input

5

....X

.O...

...X.

Output

1

Input

5

.....

.O...

.....

Output

2

Input

3

...

...

..O

Output

4

题意:有一个3*n的矩形,除其中标为X或O的点之外的点用1*2的矩形覆盖,求有多少种发难.要求标为O的点周围存在可以移动的1*2矩形.

思路:位DP?还是状态压缩?我不太清楚这分类.f[i][j]表示填充到第i列,状态为j的方案数.其中j∈[0,7],其二进制表示第k行是否被占用.因为O点周围必须有能移动的小矩形,所以枚举小矩形的位置,分别有可能在左右或下.然后用容斥原理排除多余解.

#include<stdio.h>
#include<string.h>
const __int64 MOD=;
__int64 dp[][];
char ch[][];
int N;
__int64 DP()
{
    memset(dp,,sizeof(dp));
    dp[][]=;
    for (int i=;i<N;i++)
     for (int j=;j<;j++)
     {
         bool v[];
         for (int k=;k<;k++)
        {
            v[k]=j >> k &  ^ ;
            if (ch[k][i]=='X')
            {
                if (v[k]==false) dp[i][j]=;
                v[k]=false;
            }
        }
        if (dp[i][j]==) continue;
        if (v[])
        {
            if (v[])
            {
                if (v[])
                {
                    (dp[i+][]+=dp[i][j])%=MOD;
                    (dp[i+][]+=dp[i][j])%=MOD;
                    (dp[i+][]+=dp[i][j])%=MOD;
                }
                else
                {
                    (dp[i+][]+=dp[i][j])%=MOD;
                    (dp[i+][]+=dp[i][j])%=MOD;
                }
            }
            else
            {
                if (v[])
                {
                    (dp[i+][]+=dp[i][j])%=MOD;
                }
                else
                {
                    (dp[i+][]+=dp[i][j])%=MOD;
                }
            }
        }
        else
        {
            if (v[])
            {
                if (v[])
                {
                    (dp[i+][]+=dp[i][j])%=MOD;
                    (dp[i+][]+=dp[i][j])%=MOD;
                }
                else
                {
                    (dp[i+][]+=dp[i][j])%=MOD;
                }
            }
            else
            {
                if (v[])
                {
                    (dp[i+][]+=dp[i][j])%=MOD;
                }
                else
                {
                    (dp[i+][]+=dp[i][j])%=MOD;
                }
            }
        }
     }
     return (dp[N][]);
}
int main()
{
    int sx,sy;
    while (scanf("%d",&N)!=EOF)
    {
        for (int i=;i<;i++) scanf("%s",ch[i]);
        for (int i=;i<;i++)
         for (int j=;j<N;j++)
         if (ch[i][j]=='O')
         {
             sx=i;
             sy=j;
             ch[i][j]='X';
         }
        if (sx==)
        {
            for (int i=;i<N;i++)
            {
                char tmp=ch[][i];
                ch[][i]=ch[][i];
                ch[][i]=tmp;
            }
            sx=;
        }
        ch[][N]=ch[][N]=ch[][N]='X';
        bool f1=false,f2=false,f3=false;
        __int64 ans=;
        if (sy> && ch[sx][sy-]=='.' && ch[sx][sy-]=='.')
        {
            f1=true;
            ch[sx][sy-]=ch[sx][sy-]='X';
            ans=(ans+DP())%MOD;
            ch[sx][sy-]=ch[sx][sy-]='.';
        }
        if (sy<N- && ch[sx][sy+]=='.' && ch[sx][sy+]=='.')
        {
            f2=true;
            ch[sx][sy+]=ch[sx][sy+]='X';
            ans=(ans+DP())%MOD;
            ch[sx][sy+]=ch[sx][sy+]='.';
        }
        if (sx== && ch[sx+][sy]=='.' && ch[sx+][sy]=='.')
        {
            f3=true;
            ch[sx+][sy]=ch[sx+][sy]='X';
            ans=(ans+DP())%MOD;
            ch[sx+][sy]=ch[sx+][sy]='.';
        }
        if (f1 && f2)
        {
            ch[sx][sy-]=ch[sx][sy-]=ch[sx][sy+]=ch[sx][sy+]='X';
            ans=(ans+MOD-DP())%MOD;
            ch[sx][sy-]=ch[sx][sy-]=ch[sx][sy+]=ch[sx][sy+]='.';
        }
        if (f1 && f3)
        {
            ch[sx][sy-]=ch[sx][sy-]=ch[sx+][sy]=ch[sx+][sy]='X';
            ans=(ans+MOD-DP())%MOD;
            ch[sx][sy-]=ch[sx][sy-]=ch[sx+][sy]=ch[sx+][sy]='.';
        }
        if (f2 && f3)
        {
            ch[sx][sy+]=ch[sx][sy+]=ch[sx+][sy]=ch[sx+][sy]='X';
            ans=(ans+MOD-DP())%MOD;
            ch[sx][sy+]=ch[sx][sy+]=ch[sx+][sy]=ch[sx+][sy]='.';
        }
        if (f1 && f2 && f3)
        {
            ch[sx][sy-]=ch[sx][sy-]=ch[sx][sy+]=ch[sx][sy+]=ch[sx+][sy]=ch[sx+][sy]='X';
            ans=(ans+DP())%MOD;
            ch[sx][sy-]=ch[sx][sy-]=ch[sx][sy+]=ch[sx][sy+]=ch[sx+][sy]=ch[sx+][sy]='.';
        }
        printf("%I64d\n",ans);
    }
    return ;
}

E. Xenia and Tree

time limit per test

5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Xenia the programmer has a tree consisting of n nodes. We will consider the tree nodes indexed from 1 to n. We will also consider the first node to be initially painted red, and the other nodes — to be painted blue.

The distance between two tree nodes v and u is the number of edges in the shortest path between v and u.

Xenia needs to learn how to quickly execute queries of two types:

1. paint a specified blue node in red;
2. calculate which red node is the closest to the given one and print the shortest distance to the closest red node.

Your task is to write a program which will execute the described queries.

Input

The first line contains two integers n and m (2 ≤ n ≤ 10⁵, 1 ≤ m ≤ 10⁵) — the number of nodes in the tree and the number of queries. Next n - 1 lines contain the tree edges, the i-th line contains a pair of integers a_i, b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i) — an edge of the tree.

Next m lines contain queries. Each query is specified as a pair of integers t_i, v_i (1 ≤ t_i ≤ 2, 1 ≤ v_i ≤ n). If t_i = 1, then as a reply to the query we need to paint a blue node v_i in red. If t_i = 2, then we should reply to the query by printing the shortest distance from some red node to node v_i.

It is guaranteed that the given graph is a tree and that all queries are correct.

Output

For each second type query print the reply in a single line.

Sample test(s)

Input

5 4
1 2
2 3
2 4
4 5
2 1
2 5
1 2
2 5

Output

0

3

2

题意:有一棵树,初始时编号为1的节点是红色.接下来进行m步操作,将编号为x的节点染成红色或求出x离最近红点的距离.

思路:这题麻烦之处在于数据量太大,一开始我想不断维护一个dis数组,结果惨遭TLE.于是采用另一种方法,用O(1)染色,然后每次查询时bfs一次.这两种都是O(n)级别的更新,不过很明确的一点是红点越多bfs退出的越快.当红点比较多时,每次维护dis数组需要更新的点数为原点延伸出的全蓝路径唱的一半的和,在红点很多时这个数值也可能很大.相比起来此时每次bfs找最短路所经过的节点就会少一点从而可以设一个阈值,当红点数小于这个值时维护dis数组,否则bfs查询最近红点.

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
struct node
{
    int x,step,pre;
};
int d[];
vector<int> G[];
bool col[];
void bfs(int s)
{
    queue<int> q;
    while (!q.empty()) q.pop();
    q.push(s);
    d[s]=;
    col[s]=true;
    while (!q.empty())
    {
        int x=q.front();
        q.pop();
        for (int i=;i<G[x].size();i++)
        {
            int v=G[x][i];
            if (d[v]>d[x]+)
            {
                d[v]=d[x]+;
                q.push(v);
            }
        }
    }
}
int BFS(int s)
{
    if (col[s]) return ;
    queue<node> q;
    while (!q.empty()) q.pop();
    node S;
    S.x=s;
    S.step=;
    S.pre=;
    q.push(S);
    while (!q.empty())
    {
        S=q.front();
        q.pop();
        for (int i=;i<G[S.x].size();i++)
        {
            int v=G[S.x][i];
            if (v==S.pre) continue;
            if (col[v]) return S.step+;
            else
            {
                node tmp;
                tmp.x=v;
                tmp.step=S.step+;
                tmp.pre=S.x;
                q.push(tmp);
            }
        }
    }
}
int main()
{
    int N,M;
    while (scanf("%d%d",&N,&M)!=EOF)
    {
        for (int i=;i<=N;i++) G[i].clear();
        for (int i=;i<N;i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            G[x].push_back(y);
            G[y].push_back(x);
        }
        for (int i=;i<=N;i++) d[i]=0x7FFFFFFF;
        memset(col,false,sizeof(col));
        col[]=true;
        bfs();
        int cnt=;
        for (int i=;i<=M;i++)
        {
            int Q,P;
            scanf("%d%d",&Q,&P);
            if (Q==)
            {
                cnt++;
                if (cnt<) bfs(P);
                else col[P]=true;
            }
            else
            {
                if (cnt<) printf("%d\n",d[P]);
                else printf("%d\n",BFS(P));
            }
        }
    }
    return ;
}