[luoguP2016] 战略游戏（DP）

传送门

f[i][0]表示不选当前节点，当前节点的所有儿子节点都选
f[i][1]表示选当前节点，儿子节点可选可不选

#include <cstdio>
#include <cstring>
#include <iostream>
#define N 1501
#define min(x, y) ((x) < (y) ? (x) : (y))

int n, cnt;
int head[N], to[N << 1], next[N << 1], f[N][2];
bool vis[N];

//f[i][0]表示不选当前节点，当前节点的所有儿子节点都选
//f[i][1]表示选当前节点，儿子节点可选可不选 

inline int read()
{
	int x = 0, f = 1;
	char ch = getchar();
	for(; !isdigit(ch); ch = getchar()) if(ch == ‘-‘) f = -1;
	for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - ‘0‘;
	return x * f;
}

inline void add(int x, int y)
{
	to[cnt] = y;
	next[cnt] = head[x];
	head[x] = cnt++;
}

inline void dfs(int u)
{
	int i, v;
	f[u][1] = vis[u] = 1;
	for(i = head[u]; i ^ -1; i = next[i])
	{
		v = to[i];
		if(!vis[v])
		{
			dfs(v);
			f[u][0] += f[v][1];
			f[u][1] += min(f[v][0], f[v][1]);
		}
	}
}

int main()
{
	int i, j, k, x, y;
	n = read();
	memset(head, -1, sizeof(head));
	for(i = 1; i <= n; i++)
	{
		x = read() + 1;
		k = read();
		for(j = 1; j <= k; j++)
		{
			y = read() + 1;
			add(x, y);
			add(y, x);
		}
	}
	dfs(1);
	printf("%d\n", min(f[1][0], f[1][1]));
	return 0;
}