poj1463 Strategic game 树形dp
2017-09-16 22:39
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题目:
http://poj.org/problem?id=1463题意:
给定一个有n个点的树结构,从中选取一些点,使得任意一条边都有端点在选取的点集中,即求树形的最小点覆盖思路:
定义dp[i][0]为不选取i时覆盖以i为根的子树上的所有边时选取的最小点集,dp[i][1]为选取i时覆盖以i为根的子树上的所有边时选取的最小点集,设j为i的子节点,那么显然有dp[i][0]+=dp[j][1]
dp[i][1]+=min(dp[j][0],dp[j][1])
这里初始化dp[i][1]=1
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int N = 2000 + 10, INF = 0x3f3f3f3f; struct edge { int to, next; }g[N*2]; int cnt, head ; int dp [2]; void init() { cnt = 0; memset(head, -1, sizeof head); memset(dp, 0, sizeof dp); } void add_edge(int v, int u) { g[cnt].to = u, g[cnt].next = head[v], head[v] = cnt++; } void dfs(int v, int fa) { for(int i = head[v]; ~i; i = g[i].next) { int u = g[i].to; if(u == fa) continue; dfs(u, v); dp[v][0] += dp[u][1]; dp[v][1] += min(dp[u][0], dp[u][1]); } } int main() { int n; while(~ scanf("%d", &n)) { init(); int x, y, m; for(int i = 1; i <= n; i++) dp[i][1] = 1; for(int i = 1; i <= n; i++) { scanf("%d:(%d)", &x, &m); for(int j = 1; j <= m; j++) { scanf("%d", &y); add_edge(x+1, y+1); add_edge(y+1, x+1); } } dfs(1, 0); printf("%d\n", min(dp[1][0], dp[1][1])); } return 0; }
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