【SRM 716 DIV 1 A】 ConstructLCS

Problem Statement

A string S is a subsequence of a string T if we can obtain S from T by erasing some (possibly all or none) of its characters. For example, “000” is a subsequence of “01010”. The longest common subsequence (LCS) of two strings A and B is a string C that is a subsequence of each of them and has the largest length among all strings with this property. Let f(A,B) be the length of the LCS of strings A and B. For example, we have f(“101”, “111000”) = 2, f(“101”, “110011”) = 3, and f(“00”, “1111”) = 0. You are given three small positive integers ab, bc, and ca. Please find three strings A, B, C such that:

Each of the strings contains only the characters ‘0’ and ‘1’.

The length of each string is between 1 and 1,000, inclusive.

f(A, B) = ab

f(B, C) = bc

f(C, A) = ca

Return a string formed as follows: A + ” ” + B + ” ” + C. (I.e., the returned string should contain the three strings A, B, C, separated by single spaces.) You may assume that a solution always exist. If there are multiple solutions you may return any of them.

Definition

Class:

ConstructLCS

Method:

construct

Parameters:

int, int, int

Returns:

string

Method signature:

string construct(int ab, int bc, int ca)

(be sure your method is public)

Limits

Time limit (s):

2.000

Memory limit (MB):

256

Stack limit (MB):

256

Constraints

ab will be between 1 and 50, inclusive.

bc will be between 1 and 50, inclusive.

ca will be between 1 and 50, inclusive.

Examples

0)

3

4

2

Returns: “101 1010101 1111”

The returned string corresponds to the following solution:

A = “1111”

B = “101”

C = “1010101”

We can easily verify that the only LCS of A and B is “11”, the only LCS of B and C is “101”, and the only LCS of C and A is “1111”.

1)

7

4

4

Returns: “10101010 1010101 1011”

There are other solutions like: A = “1110000”, B = “1110000”, C = “0000”.

2)

8

7

8

Returns: “110101001011 010101101 10101010”

3)

8

6

7

Returns: “110101010 10101010 1111010”

4)

15

17

19

Returns: “000100101101111011000 11110111010011101010 100100001010101001010101000011111”

5)

50

50

50

Returns:

“11111111111111111111111111111111111111111111111111 11111111111111111111111111111111111111111111111111 11111111111111111111111111111111111111111111111111”

【题目链接】:

【题意】



让你构造出3个只包含数字0和1的字符串a,b,c;

给你3个数字ab,bc,ca

表示所要求的a和b的LCS长度为ab,b和C的LCS长度为….

【题解】



找出最大的两个数字;

它所代表的两条边,必然有同时连向同一个点x;

“a、b、c分别代表3个点”

设两条边的边权从大到小分别为a1,a2

则令x+=a1个’1’;

x+=a2个’0’;

然后令a1的另一端的点所代表的字符串为a1个’1’

同时a2的另一端的点所代表的字符串为a2个’0’

然后a3的话,可以不断的在只含‘0’的那个字符串的开头一直把‘0’换成1;直到满足a3为止.



【Number Of WA】



0



【反思】



想得太慢了,没来得及交上去QAQ;

狂跌100+rating;

下次要打div2了:(



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ms(x,y) memset(x,y,sizeof x)
#define Open() freopen("F:\\rush.txt","r",stdin)
#define Close() ios::sync_with_stdio(0)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int N = 110;
//head

struct abc{
    int idx1,idx2,num;
};

abc a[4];
int b[4];
vector <pii> g[4];
string s[4];

class ConstructLCS
{
    public:
        string construct(int ab, int bc, int ca)
        {
            rep1(i,1,3) s[i]="";
            a[1].idx1 = 1,a[1].idx2 = 2,a[1].num = ab;
            a[2].idx1 = 2,a[2].idx2 = 3,a[2].num = bc;
            a[3].idx1 = 1,a[3].idx2 = 3,a[3].num = ca;
            sort(a+1,a+1+3,[&] ( abc a ,abc b) {return a.num > b.num;});
            int ma = 0;
            rep1(i,1,2){
                int x = a[i].idx1,y = a[i].idx2;
                b[a[i].idx1]++,b[a[i].idx2]++;
                ma = max(ma,a[i].num);
                g[x].pb(mp(y,a[i].num)),g[y].pb(mp(x,a[i].num));
            }

            int mid = 1;
            rep1(i,1,3){
                if (b[i]>1){
                    mid = i;
                    rep1(j,1,max(g[i][0].se,g[i][1].se))
                        s[i]+='1';
                    rep1(j,1,min(g[i][0].se,g[i][1].se))
                        s[i]+='0';
                    break;
                }
            }
            int bigger = 0,smaller = 0;
            rep1(i,1,3){
                if (mid==i) continue;
                if (bigger==0 && ma==g[i][0].se){
                    rep1(j,1,ma)
                        s[i]+='1';
                    bigger = i;
                }
                else{
                    rep1(j,1,g[i][0].se)
                        s[i]+='0';
                    smaller = i;
                }
            }
            int t = a[3].num;
            int now = 0;
            while (t--){
                s[smaller][now++] = '1';
            }
            return s[1] + ' ' + s[2] + ' ' + s[3];
        }
};