SPOJ AMR12B 720
这个题应该是个优先队列的模版题 当时比赛的时候没时间做先贴一下大神的代码好好学习学习
Time Limit:2000MS Memory Limit:0KB 64bit IO Format:%lld & %llu
Description
'We fought far under the living earth, where time is not counted. Ever he clutched me, and ever I hewed him, till at last he fled into dark tunnels. Ever up now we went, until we came to the Endless Stair. Out he sprang, and even as I came behind, he burst into new flame. Those that looked up from afar thought that the mountain was crowned with storm. Thunder they heard, and lightning, they said, smote upon Celebdil, and leaped back broken into tongues of fire.' - Gandalf, describing his fight against the Balrog.
Although Gandalf would not go into the details of his battle, they can be summarized into the following simplified form: both Gandalf and the Balrog have a set of N attacks they can use (spells, swords, brute-force strength etc.). These attacks are numbered from 1 to N in increasing order of Power. When each has chosen an attack, in general, the one with the higher power wins. However, there are a few ("M") anomalous pairs of attacks, in which the lesser-powered attack wins.
Initially, Gandalf picks an attack. Then the Balrog counters it with one of the remaining attacks. If the Balrog's counter does not defeat Gandalf's, then we say Gandalf receives a score of 2. If however it does, then Gandalf has exactly one more opportunity to pick an attack that will defeat the Balrog's. If he manages to defeat him now, his score will be 1, whereas if he is still unable to defeat him, his score will be 0.
Your task is to determine, given N and the M anomalous pairs of attacks, what will be Gandalf's score, given that both play optimally. Further, in case Gandalf gets a score of 2, you must also determine which attack he could have chosen as his first choice.
Note 1: The Balrog can choose only one attack, whereas Gandalf can choose upto two.
Note 2: The relation A defeats B is not transitive within the attacks. For example, attack A can defeat attack B, attack B can defeat attack C, and attack C can defeat attack A.
Note 3: Between any two attacks A and B, either attack A defeats attack B or attack B defeats attack A.
Input (STDIN):
The first line will consist of the integer T, the number of test-cases.
Each test case begins with a single line containing two integers N and M.
This is followed by M lines consisting of 2 integers each x and y, denoting that x and y are an anomalous pair.
Output (STDOUT):
For each test-case, output a single line either
2 A, if Gandalf can defeat any attack the Balrog chooses if he picks attack A,
1, if Gandalf can choose an attack such that even if the Balrog chooses an attack to defeat him, he can choose an attack to defeat the Balrog's chosen card,
0, if none of the above two options are possible for all possible choices of Gandalf's attack(s).
Between successive test cases, there should not be any blank lines in the output.
Constraints:
1 <= T <= 15
3 <= N <= 1,000,000
0 <= M <= min(N(N-1)/2, 300,000)
1 <= x < y <= N for all the anomalous pairs (x,y)
The sum of M over all test-cases will not exceed 300,000.
Sample Input:
2
3 0
3 1
1 3
Sample Output:
2 3
1
Notes/Explanation of Sample Input:
In the first case, attack 3 can beat both attacks 1 and 2. So Gandalf just chooses attack 3.
In the second case, attack 1 beats 3 which beats 2 which beats 1. No matter which attack Gandalf chooses, the Balrog can pick the one which defeats his, but then he can pick the remaining attack and defeat the Balrog's.
#include <iostream> using namespace std; int array[]; int main()
{
int t, n, m, x, y;
cin >> t; while (t--)
{
cin >> n >> m; for (int i = ; i <= n; i++)
{
array[i] = n - i;
} for (int i = ; i < m; i++)
{
cin >> x >> y;
array[x]--;
array[y]++;
} int ok = ;
for (int i = ; i <= n; i++)
{
if (array[i] == )
{
ok = i;
break;
}
} if (ok)
{
cout << << ' ' << ok << endl;
}
else
{
cout << << endl;
}
}
return ;
}
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<string>
#include<iostream>
#include<vector>
#define N 1001000 using namespace std; int in[N];
int out[N];
int n,m;
int main()
{
int cas;
scanf("%d",&cas);
while (cas--){
scanf("%d%d",&n,&m);
memset(in,,sizeof(in));
memset(out,,sizeof(out)); for (int i=;i<=m;i++){
int x,y;
scanf("%d%d",&x,&y);
in[y]++;
out[x]++;
}
int i = n;
while (i>= && !(in[i]== && out[i]>=(n - i))){
i--; }
if (i>=){
printf("2 %d\n",i);
}else{
if (m == (n-)*n /)
printf("0\n");
else
printf("1\n");
}
}
return ;
}
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <cctype>
#include <climits>
#include <ctime>
#include <vector>
#include <set>
#include <stack>
#include <sstream>
#include <iomanip>
#define MAX 1000010
#define CLR(arr,val) memset(arr,val,sizeof(arr)) using namespace std; int cards[MAX], defeated[MAX]; int main()
{
std::ios::sync_with_stdio(false);
#ifndef ONLINE_JUDGE
freopen( "in.txt", "r", stdin );
//freopen( "out.txt", "w", stdout );
clock_t program_start, program_end;
program_start = clock();
#endif
int T, N, M, x, y, max_card;
cin >> T;
while ( T-- )
{
cin >> N >> M;
for ( int i = ; i <= N; ++i )
{
cards[i] = ;
defeated[i] = ;
}
for ( int i = ; i < M; ++i )
{
cin >> x >> y;
cards[x]++; defeated[y]++;
}
max_card = -;
for ( int i = N; i >= ; --i )
if ( cards[i] == N - i && defeated[i] == )
{
max_card = i;
break;
}
if ( max_card != - )
cout << "2 " << max_card << endl;
else
cout << "" << endl;
} #ifndef ONLINE_JUDGE
program_end = clock();
cerr << "Time consumed: " << endl << ( program_end - program_start ) << " MS" << endl;
#endif
}
SPOJ AMR12B 720的更多相关文章
- BZOJ 2588: Spoj 10628. Count on a tree [树上主席树]
2588: Spoj 10628. Count on a tree Time Limit: 12 Sec Memory Limit: 128 MBSubmit: 5217 Solved: 1233 ...
- SPOJ DQUERY D-query(主席树)
题目 Source http://www.spoj.com/problems/DQUERY/en/ Description Given a sequence of n numbers a1, a2, ...
- SPOJ GSS3 Can you answer these queries III[线段树]
SPOJ - GSS3 Can you answer these queries III Description You are given a sequence A of N (N <= 50 ...
- 【填坑向】spoj COT/bzoj2588 Count on a tree
这题是学主席树的时候就想写的,,, 但是当时没写(懒) 现在来填坑 = =日常调半天lca(考虑以后背板) 主席树还是蛮好写的,但是代码出现重复,不太好,导致调试的时候心里没底(虽然事实证明主席树部分 ...
- SPOJ bsubstr
题目大意:给你一个长度为n的字符串,求出所有不同长度的字符串出现的最大次数. n<=250000 如:abaaa 输出: 4 2 1 1 1 spoj上的时限卡的太严,必须使用O(N)的算法那才 ...
- 【SPOJ 7258】Lexicographical Substring Search
http://www.spoj.com/problems/SUBLEX/ 好难啊. 建出后缀自动机,然后在后缀自动机的每个状态上记录通过这个状态能走到的不同子串的数量.该状态能走到的所有状态的f值的和 ...
- 【SPOJ 1812】Longest Common Substring II
http://www.spoj.com/problems/LCS2/ 这道题想了好久. 做法是对第一个串建后缀自动机,然后用后面的串去匹配它,并在走过的状态上记录走到这个状态时的最长距离.每匹配完一个 ...
- 【SPOJ 8222】Substrings
http://www.spoj.com/problems/NSUBSTR/ clj课件里的例题 用结构体+指针写完模板后发现要访问所有的节点,改成数组会更方便些..于是改成了数组... 这道题重点是求 ...
- SPOJ GSS2 Can you answer these queries II
Time Limit: 1000MS Memory Limit: 1572864KB 64bit IO Format: %lld & %llu Description Being a ...
随机推荐
- django+pymysql搭建一个管理系统(一)
django+pymysql搭建一个管理系统(一) 后续进行代码更新,优化 一.程序架构 二.mysql表单创建 zouye库:存信息相关的 #班级表 create table classes( ci ...
- SqlServer 附加数据库出错
方法一 找到要添加数据库的.mdf文件,点击右键,选择属性 在属性页面点击安全,选择Authenticated Users,单击编辑 Authenticated Users权限中选择完全控制,点击确定 ...
- ES6入门六:class的基本语法、继承、私有与静态属性、修饰器
基本语法 继承 私有属性与方法.静态属性与方法 修饰器(Decorator) 一.基本语法 class Grammar{ constructor(name,age){ //定义对象自身的方法和属性 t ...
- Oracle---PL/SQL的学习
PL/SQL程序 一.定义 declare 说明部分 begin 语句序列(DML语句) exception 例外处理语句 end; 二. 变量和常量说明 a) 说明变量(char,varchar2, ...
- extension(类扩展)和 category(类别)
extension(类扩展) 简单来说,extension在.m文件中添加,所以其权限为private,所以只能拿到源码的类添加extension.另外extension是编译时决议,和interfa ...
- printf颜色
格式 printf("\033[?m%s\033[0m", str); 多个属性以:分隔 属性: \033[0m:关闭所有属性 \033[1m:设置高亮度 \033[4m:下划线 ...
- Java,JavaScript和ABAP通过代码取得当前代码的调用栈Callstack
Java StackTraceElement stack[] = Thread.currentThread().getStackTrace(); System.out.println("Ca ...
- Linux软链接创建及删除
1.创建软链接 具体用法是:ln -s [源文件] [软链接文件]. [root@localhost folder]# pwd /tmp/folder [root@localhost fol ...
- go语言入门(10)并发编程
1,概述 1.1并发和并行 并行(parallel):指在同一时刻,有多条指令在多个处理器上同时执行. 并发(concurrency):指在同一时刻只能有一条指令执行,但多个进程指令被快速的轮换执行, ...
- 如何处理不能新建word、excel、PPT的情况?
Office系列办公软件是大家都非常喜欢使用的软件,但是有些朋友反映在使用电脑时,在桌面右键菜单新建选项里没有Word.Excel或PPT,非常的耽误工作. 下面就为大家介绍一下桌面右键菜单新建选项里 ...