Dominator UVA - 11902

题目链接

题意

如果有一个点 $b$ ，从 $1$ （我喜欢从 $1$ 开始编号）到 $b$ 的路径（不一定是最短路径）必须要经过点 $a$ ，那么答案数组 $ans[a][b]=Y$ ，否则 $ans[a][b]=N$
输出 $ans$ 数组。

思路

如果把一个点 $a$ 删掉后有几个原来能从 $1$ 走到的点 $y_1,y_2,...,y_k$ 不能从 $1$ 走到了，那么从 $1$ 到这几个点的路径一定要经过点 $a$ 即 $ans[a][y_1]=ans[a][y_2]=...=ans[a][y_k]=Y$

代码

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 110;
int test_case = 1;
int n;
bool g[N][N],backup_g[N][N];
bool vis[N],backup_vis[N];
bool ans[N][N];
void dfs (int x) {
	if (vis[x]) return ;
	vis[x] = true;
	for (int i = 1;i <= n;i++) {
		if (g[x][i] && !vis[i]) dfs (i);
	}
}
int main () {
	int T;
	cin >> T;
	while (T--) {
		cin >> n;
		for (int i = 1;i <= n;i++) {
			for (int j = 1;j <= n;j++) {
				cin >> g[i][j];
				backup_g[i][j] = g[i][j];
			}
		}
		memset (vis,false,sizeof (vis));
		dfs (1);
		memcpy (backup_vis,vis,sizeof (vis));
		memset (ans,false,sizeof (ans));
		for (int i = 1;i <= n;i++) {
			if (vis[i]) ans[1][i] = ans[i][i] = true;
		}
		for (int i = 2;i <= n;i++) {
			memset (g[i],false,sizeof (g[i]));
			memset (vis,false,sizeof (vis));
			dfs (1);
			for (int j = 1;j <= n;j++) {
				if (!vis[j] && backup_vis[j]) ans[i][j] = true;
			}
			memcpy (g[i],backup_g[i],sizeof (backup_g[i]));
		}
		cout << "Case " << test_case++ << ":" << endl;
		for (int i = 1;i <= 2 * n + 1;i++) {
			if (i & 1) {
				cout << "+";
				for (int j = 2;j <= 2 * n;j++) cout << "-";
				cout << "+";
			}
			else {
				for (int j = 1;j <= 2 * n + 1;j++) {
					if (j & 1) cout << "|";
					else {
						if (ans[i / 2][j / 2]) cout << "Y";
						else cout << "N";
					}
				}
			}
			cout << endl;
		}
	}
    return 0;
}