来源:POJ1463
树形dp
代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
//原问题:在一棵树中我需要多少人来守卫
//子问题:在子树中需要多少人来守卫
//那问题就是我们考虑树根放还是不放
//对于一条边而言,我们可以是上和下或者同时来监督
//从而,对于一个节点,我们就有三个情况:
//dp[rt][0]表示这一点放了,dp[rt][1]表示这一点没有、
//dp[rt][0]=1+sum(min(dp[son][0],dp[son][1]))
//dp[rt][1]=sum(dp[son][1])
const int MAXN=1500+10;
int n;
int d;
vector<int>tree[MAXN];
int sum[MAXN];
int dp[MAXN][2];
bool vis[MAXN];
bool son[MAXN];
int fa;
void dfs(int rt){
vis[rt]=1;
dp[rt][1]=1;
dp[rt][0]=0;
/*cout<<"rt: "<<rt<<endl;
cout<<rt<<" son:"<<endl;*/
for(int i=0;i<sum[rt];i++){
//cout<<tree[rt][i]<<":"<<endl;
if(vis[tree[rt][i]]==0){
dfs(tree[rt][i]);
dp[rt][0]+=dp[tree[rt][i]][1];
dp[rt][1]+=min(dp[tree[rt][i]][0],dp[tree[rt][i]][1]);
//cout<<rt<<" "<<tree[rt][i]<<" "<<dp[tree[rt][i]][0]<<" "<<dp[tree[rt][i]][1]<<endl;
}
}
}
int main(){
char shitc;
while(scanf("%d",&n)!=EOF){
memset(vis,0,sizeof(vis));
memset(son,0,sizeof(son));
for(int i=0;i<n;i++) tree[i].clear();
for(int i=0;i<n;i++){
scanf("%d",&fa);
//cout<<"fa: "<<fa<<endl;
scanf("%c",&shitc);
//cout<<"shitc: "<<shitc<<endl;
scanf("%c",&shitc);
//cout<<"shitc: "<<shitc<<endl;
scanf("%d",&sum[fa]);
//cout<<"sum: "<<sum[fa]<<endl;
scanf("%c",&shitc);
//cout<<"shitc: "<<shitc<<endl;
for(int j=0;j<sum[fa];j++){
scanf("%d",&d);
//cout<<"son "<<d<<endl;
tree[fa].push_back(d);
son[d]=1;
}
//getchar();
}
//cout<<tree[1][0]<<"Dfdsds"<<endl;
for(int i=0;i<n;i++){
if(!son[i]){
//cout<<"root "<<i<<endl;
dfs(i);
cout<<min(dp[i][0],dp[i][1])<<endl;
break;
}
}
}
return 0;
}