Strategic game POJ - 1463

题目链接

依旧是树形dp啦，一样的找根节点然后自下而上更新即可

设\(dp_{i,0}\)表示第i个不设，\(dp_{i,1}\)表示第i个设一个，容易得到其状态转移

\(dp_{i,0} = \sum{dp_{j,1}}(j为i的儿子节点)\)

\(dp_{i,1} = 1 + \sum{min(dp_{j,0}, dp_{j,1})}(j为i的儿子节点)\)

#include<iostream>
#include<vector>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;

const int maxn = 1507;
vector<int> son[maxn];
int in[maxn], dp[maxn][2];

void dfs(int fa) {
    dp[fa][0] = 0;
    dp[fa][1] = 1;
    for(int i = 0; i < son[fa].size(); ++i) {
        int v = son[fa][i];
        dfs(v);
        dp[fa][1] += min(dp[v][0], dp[v][1]);
        dp[fa][0] += dp[v][1];
    }
}

void run_case() {
    int n;
    while(cin >> n) {
        memset(dp, 0, sizeof(dp)), memset(in, 0, sizeof(in));
        for(int i = 0; i < n; ++i) {
            int fa, num;
            scanf("%d:(%d)", &fa, &num);
            son[fa].clear();
            for(int j = 0; j < num; ++j) {
                int v;
                scanf("%d", &v);
                in[v]++;
                son[fa].push_back(v);
            }
        }
        for(int i = 0; i < n; ++i)
            if(!in[i]) {
                dfs(i);
                cout << min(dp[i][0], dp[i][1]) << "\n";
                break;
            }
    }

}

int main() {
    //ios::sync_with_stdio(false), cout.tie(0);
    cout.flags(ios::fixed);cout.precision(10);
    run_case();
    //cout.flush();
    return 0;
}