标签：poj   acm   treedp   

Description

Bob enjoys playing computer games, especially strategic games, but sometimes he cannot find the solution fast enough and then he is very sad. Now he has the following problem. He must defend a medieval city, the roads of which form a tree. He has to put the minimum number of soldiers on the nodes so that they can observe all the edges. Can you help him?

Your program should find the minimum number of soldiers that Bob has to put for a given tree.

For example for the tree:

the solution is one soldier ( at the node 1).

Input

Output

Sample Input

4
0:(1) 1
1:(2) 2 3
2:(0)
3:(0)
5
3:(3) 1 4 2
1:(1) 0
2:(0)
0:(0)
4:(0)

Sample Output

1
2

还是对树地守卫问题，不过这题是守卫所有的边。节点可以安排士兵，其能守卫相邻的边。求用最少的士兵，守卫所有的边。与皇宫守卫不同。这里只需要看节点是否安排守卫。如果不安排，则其所有子节点均要安排，如果安排守卫，则其子节点状态随意。

//dp[t][0/1] : 根节点为t的子树(0:根节点不安排守卫1:安排)的所有边被守卫的情况下的最小安排士兵数量。

dp[t][0] = sum(dp[ti][1]);
dp[t][1] = sum(min(dp[ti][0], dp[ti][1]));

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

const int INF = 999999999;

int n;
std::vector<int > v[2005];
bool vis[2005];
int dp[2005][2];

void Tdp(int t) {
	if (v[t].size() == 0) {
		dp[t][0] = 0;
		dp[t][1] = 1;
		return ;
	}
	for (int i=0; i<v[t].size(); i++) {
		Tdp(v[t][i]);
		dp[t][0] += dp[v[t][i]][1];
		dp[t][1] += min(dp[v[t][i]][0], dp[v[t][i]][1]);
	}
	dp[t][1]++;
}

void init() {
	memset(dp, 0, sizeof(dp));
	memset(vis, false, sizeof(vis));
	for (int i=0; i<1505; i++) {
		v[i].clear();
	}
}

int main () {
	while (scanf ("%d",&n)!=EOF) {
		init();
		for (int i=0; i<n; i++) {
			int a, b, c;
			scanf ("%d:(%d)", &a, &b);
			for (int j=0; j<b; j++) {
				scanf ("%d", &c);
				v[a].push_back(c);
				vis[c] = true;
			}
		}		int root;
		for (int i=0; i<n; i++) {
			if (!vis[i]) {
				root = i;
				break;
			}
		}
		Tdp(root);
		cout << min(dp[root][0], dp[root][1]) << endl;
	}
	return 0;
}

poj 1463 Strategic game 树状dp

标签：poj   acm   treedp   

原文地址：http://blog.csdn.net/xuelanghu407/article/details/43700619