本文主要是介绍uva 10859,希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!
刘书例题 树形dp#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <sstream>
#include <string>
#include <cstring>
#include <algorithm>
#include <iostream>
#define maxn 200010
#define INF 0x7fffffff
#define inf 10000000
#define MOD 1000000007
#define ull unsigned long long
#define ll long long
using namespace std;vector<int> g[1010];bool vis[1010][2];
int d[1010][2], m, n;int dp(int i, int j, int fa)
{if(vis[i][j]) return d[i][j];vis[i][j] = true;int& ans = d[i][j];ans = 2000;for(int k = 0; k < (int)g[i].size(); ++ k)if(g[i][k] != fa) ans += dp(g[i][k], 1, i);if(!j && fa >= 0) ++ ans;if(j || fa < 0){int sum = 0;for(int k = 0; k < (int)g[i].size(); ++ k)if(g[i][k] != fa) sum += dp(g[i][k], 0, i);if(fa >= 0) ++ sum;ans = min(ans, sum);}return ans;
}void init()
{for(int i = 0; i < n; ++ i) g[i].clear();memset(vis, 0, sizeof(vis));
}int main()
{int t;scanf("%d", &t);while(t --){scanf("%d%d", &n, &m);init();for(int i = 0; i < m; ++ i){int a, b;scanf("%d%d", &a, &b);g[a].push_back(b);g[b].push_back(a);}int ans = 0;for(int i = 0; i < n; ++ i)if(!vis[i][0]) ans += dp(i, 0, -1);printf("%d %d %d\n", ans/2000, m-ans%2000, ans%2000);}return 0;
}
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