ZOJ 1134 Strategic Game(树形DP)
2017-08-01 22:31
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题目链接:点击打开链接
题意:一个树形图,在节点上安排士兵,每个士兵能看到与这个节点相连的边,问最少安排多少士兵能看到所有的边
很显然是树形DP,二维数组dp[i][2],dp[i][0]是以当前节点为根的子树,在树根不放士兵的情况下该子树最少安排的士兵;dp[i][1]是以当前节点为根的子树,在树根放士兵的情况下该子树最少安排的士兵
题意:一个树形图,在节点上安排士兵,每个士兵能看到与这个节点相连的边,问最少安排多少士兵能看到所有的边
很显然是树形DP,二维数组dp[i][2],dp[i][0]是以当前节点为根的子树,在树根不放士兵的情况下该子树最少安排的士兵;dp[i][1]是以当前节点为根的子树,在树根放士兵的情况下该子树最少安排的士兵
// ZOJ 1134 Strategic Game.cpp 运行/限制:150ms/1000ms #include <cstdio> #include <cstring> #include <vector> #include <algorithm> #include <iostream> using namespace std; vector<int> tree[1505]; bool book[1505]; int dp[1505][2]; void init() { for (int i = 0; i < 1505; i++) { tree[i].clear(); } memset(book, false, sizeof(book)); } int findFather() { for (int i = 0; i < 1505; i++) { if (book[i] == false) { return i; } } } int DP(int fa) { dp[fa][1] = 1; dp[fa][0] = 0; for (int i = 0; i < tree[fa].size(); i++) { int v = tree[fa][i]; DP(v); dp[fa][1] += min(dp[v][0], dp[v][1]); dp[fa][0] += dp[v][1]; } return min(dp[fa][0], dp[fa][1]); } int main(){ int n, fa; int u, v, cnt; while (scanf("%d", &n) != EOF) { init(); for (int i = 0; i < n; i++) { scanf("%d", &u); getchar();getchar(); scanf("%d", &cnt); getchar(); for (int j = 0; j < cnt; j++) { scanf("%d", &v); tree[u].push_back(v); book[v] = true; } } fa = findFather(); printf("%d\n",DP(fa)); } return 0; }
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