TZOJ--5449: King of the Waves（拓扑排序）

5449: King of the Waves 

时间限制(普通/Java):5000MS/15000MS     内存限制:65536KByte

描述

You are organising a king of the hill tournament, the Buenos Aires Paddleboarding Competition (BAPC), with n participants. In a king of the hill tournament, one person starts as a “king” and is then challenged by another person, the winning person becomes the new king. This is repeated until all participants have challenged exactly once (except for the starting person). In a paddleboarding match, there are no draws. The person which ends up as king, wins the tournament. Since you are the organiser, you get to choose the starting person and the order in which they challenge the king. 

Someone is offering you a substantial amount of money in case one of the participants, Henk, ends up winning the tournament. You happen to know, for any two participants x and y, which of the two would win if they were to match during the tournament. Consequently, you choose to do the unethical: you will try to rig the game. Can you find a schedule that makes Henk win the tournament?

输入

The first line contains an integer 1 ≤ n ≤ 1000, the number of participants. The participants are numbered 0, . . . , n − 1, where Henk is 0. 

• Then n lines follow, where each line has exactly n characters (not counting the newline character). These lines represent the matrix with the information of who beats who, as follows. On line i the jth character is (note that 0 ≤ i, j < n): 

– ’1’ if person i will win against person j. 

– ’0’ if person i will lose against person j. 

– ’X’ if i = j.

输出

If there exists a way to rig the game such that Henk wins, print "Yes", else "No".

样例输入

3
X10
0X1
10X

样例输出

 Yes

题目来源

BAPC 2017

 

题目链接：http://tzcoder.cn/acmhome/problemdetail.do?&method=showdetail&id=5449

 

题目大意：有N个人，给你一幅N*N的图，第i行第J列表示i和j的比赛结果，1表示赢，2表示输，问你能否寻找一种方案使得0号选手获胜。那么即寻找一种方案使得拓扑排序后0大于所有即可。（这题时间好长，应该其他方法瞎搞也可以过）

 

 

#include <bits/stdc++.h>
using namespace std;  
#define LL long long  
void closeio(){
	cin.tie(0);
   	ios::sync_with_stdio(0);
}
vector<int>g[1010];
char ma[1010][1010];
int vis[1001],n,ans;
int flag = 0;
void dfs(int v){
	vis[v] = 1;
	if(ans == n-1){
		flag = 1;
		return;
	}
	for(int i = 0 ; i < g[v].size() ; i ++){
		if(!vis[g[v][i]]){
			ans++;
			dfs(g[v][i]);
		}
	}
}
int main()
{
	scanf("%d",&n);
	flag = ans =0;
	for(int i = 0 ; i < n ; i++) g[i].clear();
	for(int i = 0 ; i < n ; i++){
		scanf("%s",ma[i]);
		for(int j = i + 1 ; j < n ; j++){
			if(ma[i][j] == '1'){
				g[i].push_back(j);
			}
			else{
				g[j].push_back(i);
			}
		}
	}
	if(n <= 1){
		puts("Yes");return 0;
	}
	memset(vis,0,sizeof(vis));
	dfs(0);
	if(flag)puts("Yes");
	else puts("No");
	return 0;
}