POJ1463 Strategic game

题解:

树形dp,最小独立集裸题

代码:

#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
using namespace std;
#define pb push_back
#define mp make_pair
#define se second
#define fs first
#define ll long long
#define CLR(x) memset(x,0,sizeof x)
#define MC(x,y) memcpy(x,y,sizeof(x))
#define SZ(x) ((int)(x).size())
#define FOR(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define INF 2097152
typedef pair<int,int> P;
const double eps=1e-9;
const int maxn=1510;
const int mod=1e9+7;

int dp[maxn][2],not_root[maxn];
vector<int> Edge[maxn];
int n,root,num;

int DP(int u,int j){
    int& ans=dp[u][j];
    if(ans>=0) return ans;
    ans=0;
    if(j) ans+=1;
    for(int i=0;i<Edge[u].size();i++){
        if(j) ans+=min(DP(Edge[u][i],0),DP(Edge[u][i],1));
        else  ans+=DP(Edge[u][i],1);
    }
    return ans;
}

int main(){
    while(~scanf("%d",&n)){
        memset(not_root,0,sizeof(not_root));
        memset(dp,-1,sizeof(dp));
        for(int i=0;i<n;i++) Edge[i].clear();
        for(int i=1;i<=n;i++){
            int u,num,v;
            scanf("%d:(%d)",&u,&num);
            for(int i=1;i<=num;i++){
                scanf("%d",&v);
                Edge[u].pb(v);
                not_root[v]=1;
            }
            for(int i=0;i<n;i++){
                if(!not_root[i]){
                    root=i;
                    break;
                }
            }
        }
        num=DP(root,0);
        memset(dp,-1,sizeof(dp));
        num=min(num,DP(root,1));
        printf("%d\n",num);
    }
    return 0;
}