POJ 1463 Strategic game 最小点覆盖，树形DP

/*应该是一道最小点覆盖问题，可以拆点用二分图做这里用树形DP来解决对叶子结点i，显然：dp[i][0]=0,dp[i][1]=1否则，dp[i][0]=sum(dp[j][1],j为i的孩子)，    dp[i][1]=sum(MIN(dp[j][0],dp[j][1]))*/#include <vector>#include <map>#include <set>#include <deque>#include <stack>#include <bitset>#include <algorithm>#include <functional>#include <numeric>#include <utility>#include <sstream>#include <iostream>#include <iomanip>#include <cstdio>#include <queue>#include <cmath>#include <cstdlib>#include <cstring>#include <ctime>#include <string>using namespace std;const int N=1600;int n;struct Edge{int v,next;}edge[2*N];int first[N],e;int DP[N][2];void addedge(int u,int v){edge[e].v=v;edge[e].next=first[u];first[u]=e++;edge[e].v=u;edge[e].next=first[v];first[v]=e++;}void init(){int x,num,y;e=0;for(int i=0;i<n;i++){first[i]=-1;DP[i][0]=0;DP[i][1]=1;}for(int i=0;i<n;i++){scanf("%d:(%d)",&x,&num);for(int j=0;j<num;j++){scanf("%d",&y);addedge(x,y);}}}int dfs(int x,int fa){int flag=1;for(int k=first[x];k!=-1;k=edge[k].next){if(edge[k].v!=fa){dfs(edge[k].v,x);DP[x][1]+=min(DP[edge[k].v][0],DP[edge[k].v][1]);DP[x][0]+=DP[edge[k].v][1];flag=0;}}}void solve(){dfs(0,-1);cout<<min(DP[0][0],DP[0][1])<<endl;}int main(){    while(scanf("%d",&n)!=EOF){init();solve();}}