解题报告 『战略游戏（树形动规）』

原题地址

qyx和lrl两位大佬表示可以用贪心秒了此题，但我觉得用树形动规也挺简单的。

代码实现如下：

#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for (register int i = a; i <= b; i++)

const int maxn = 2e3;

int n, num_edge = 0;
int head[maxn], dp[maxn][5];

struct node {
    int to, nxt;
}edge[maxn << 1];

int MIN(int a, int b) {return a < b ? a : b;}

void origin() {memset(head, -1, sizeof(head));}

int read() {
    int x = 0, flag = 0;
    char ch = ' ';
    while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
    if (ch == '-') {
        flag = 1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
        x = (x << 1) + (x << 3) + ch - '0';
        ch = getchar();
    }
    return flag ? -x : x;
}

void addedge(int from, int to) {
    edge[++num_edge].nxt = head[from];
    edge[num_edge].to = to;
    head[from] = num_edge;
}

void DP(int u, int fa) {
    dp[u][0] = 0;
    dp[u][1] = 1;
    for (register int i = head[u]; ~i; i = edge[i].nxt) {
        int v = edge[i].to;
        if (v == fa) continue;
        DP(v, u);
        dp[u][0] += dp[v][1];
        dp[u][1] += MIN(dp[v][0], dp[v][1]);
    }
}

void write(int x) {
    if (x < 0) {
        putchar('-');
        x = -x;
    }
    if (x > 9) write(x / 10);
    putchar(x % 10 + '0');
}

int main() {
    origin();
    n = read();
    rep(i, 1, n) {
        int u, v, tot;
        u = read(), tot = read();
        rep(j, 1, tot) {
            v = read();
            addedge(u, v);
            addedge(v, u);
        }
    }
    DP(0, -1);
    write(MIN(dp[0][0], dp[0][1]));
    return 0;
}

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