【洛谷P2016】战略游戏

题面

题解

树形(dp)（最大独立集）

设(f_{i,0/1})表示(dp)到第(i)个点，在这个点放了（没放）士兵的最小花费

直接转移即可。

代码

#include<cstdio>
#include<cstring>
#include<vector>
#define RG register
#define file(x) freopen(#x".in", "r", stdin);freopen(#x".out", "w", stdout);
#define clear(x, y) memset(x, y, sizeof(x));

namespace IO
{
	const int BUFSIZE = 1 << 20;
	char ibuf[BUFSIZE], *is = ibuf, *it = ibuf;
	inline char getchar() { if (is == it) it = (is = ibuf) + fread(ibuf, 1, BUFSIZE, stdin); return *is++; }
}

inline int read()
{
	int data = 0, w = 1;
	char ch = IO::getchar();
	while(ch != '-' && (ch < '0' || ch > '9')) ch = IO::getchar();
	if(ch == '-') w = -1, ch = IO::getchar();
	while(ch >= '0' && ch <= '9') data = data * 10 + (ch ^ 48), ch = IO::getchar();
	return data*w;
}

const int maxn(2000);
int f[maxn][2], n;

std::vector<int> G[maxn];
typedef std::vector<int>::iterator iter;

void dfs(int x)
{
	f[x][1] = 1; f[x][0] = 0;
	for(RG iter it = G[x].begin(); it != G[x].end(); ++it)
		dfs(*it), f[x][1] += std::min(f[*it][1], f[*it][0]), f[x][0] += f[*it][1];
}

int main()
{
#ifndef ONLINE_JUDGE
	file(cpp);
#endif
	n = read();
	for(RG int i = 1, x, num; i <= n; i++)
	{
		x = read() + 1; num = read();
		for(RG int i = 1; i <= num; i++) G[x].push_back(read() + 1);
	}
	dfs(1); printf("%d
", std::min(f[1][0], f[1][1]));
	return 0;
}