UVA 11927 - Games Are Important(sg函数)

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：求出每个结点的sg函数，然后偶数个石头结点可以不用考虑，因为对于偶数情况，总步数肯定能保证是偶数，所以只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\n", ans == 0? "Second":"First"); }return 0;}