UVA 11927

UVA 11927 - Games Are Important

题目链接

题意：给定一个有向图，结点上有一些石头，两人轮流移动石头。看最后谁不能移动就输了，问先手还后手赢

思路：求出每一个结点的sg函数，然后偶数个石头结点能够不用考虑，由于对于偶数情况，总步数肯定能保证是偶数，所以仅仅要考虑奇数情况的结点

代码：

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <vector>
using namespace std;

const int N = 1005;
int n, m, sg[N];
vector<int> g[N];

int dfs(int u) {
	if (sg[u] != -1) return sg[u];
	if (g[u].size() == 0) return sg[u] = 0;
	bool vis[N];
	memset(vis, false, sizeof(vis));
 	for (int i = 0; i < g[u].size(); i++)
		vis[dfs(g[u][i])] = true;
	for (int i = 0; ; i++)
		if (!vis[i]) return sg[u] = i;
}

int main() {
	while (~scanf("%d%d", &n, &m) && n || m) {
		int u, v;
		memset(g, 0, sizeof(g));
		memset(sg, -1, sizeof(sg));
  		while (m--) {
  			scanf("%d%d", &u, &v);
  			g[u].push_back(v);
  		}
  		for (int i = 0; i < n; i++)
    		dfs(i);
  		int ans = 0, num;
  		for (int i = 0; i < n; i++) {
  			scanf("%d", &num);
  			if (num&1)
  				ans ^= sg[i];
    	}
    	printf("%s
", ans == 0?
 "Second":"First");
 	}
	return 0;
}