uva1292(基础树形dp)

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思路：题目的意思是，游一棵树或者是森林，然后要在某些节点上放置一个警察来防卫，然后每个警察呢只能防卫到自己所在节点和相邻的节点，求所有节点都在直接或间接被防卫的时候需要的最少警察数目。

意思显然，然后就是dp了；对于当前节点的决策是选与不选，dp[i][j],表示第i个节点的是否直接放置警察；

初始化是dp[i][1]=1;

dp[u][1] += min(dp[v][1],dp[v][0]);

dp[u][0] += dp[v][1];

由于存在森林，所以要遍历一遍。

/*****************************************Author      :Crazy_AC(JamesQi)Time        :2015File Name   :*****************************************/// #pragma comment(linker, "/STACK:1024000000,1024000000")#include <iostream>#include <algorithm>#include <iomanip>#include <sstream>#include <string>#include <stack>#include <queue>#include <deque>#include <vector>#include <map>#include <set>#include <stdio.h>#include <string.h>#include <math.h>#include <stdlib.h>#include <limits.h>using namespace std;#define MEM(a,b) memset(a,b,sizeof a)typedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> ii;const int inf = 1 << 30;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;inline int Readint(){char c = getchar();while(!isdigit(c)) c = getchar();int x = 0;while(isdigit(c)){x = x * 10 + c - '0';c = getchar();}return x;}int n;vector<int> G[1510];int dp[1510][2];bool vis[1510];void dfs(int u,int fa){vis[u]=true;dp[u][1]=1;dp[u][0]=0;for (int i=0;i<G[u].size();++i){int v = G[u][i];if (v==fa || vis[v]) continue;dfs(v,u);dp[u][0] += dp[v][1];dp[u][1] += min(dp[v][0],dp[v][1]);}}int main(){// freopen("in.txt","r",stdin);// freopen("out.txt","w",stdout);while(scanf("%d",&n)!=EOF){memset(dp, 0,sizeof dp);for (int i = 0;i < n;++i) G[i].clear();for (int i = 1;i <= n;++i){int u,num;scanf("%d:(%d)",&u,&num);int v;while(num--){scanf("%d",&v);G[v].push_back(u);G[u].push_back(v);}}memset(vis, false,sizeof vis);int ans=0;for (int i = 0;i < n;++i){if (!vis[i]) {dfs(i,-1);ans += min(dp[i][1],dp[i][0]);}}printf("%d\n", ans);}return 0;}

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