uva 10600（次小生成树暴力）

题意：给你一张无向带权图，求出最小生成树和次小生成树。

思路：数据小直接爆。枚举删除最小生成树上的每一条边。

代码如下：

/**************************************************
 * Author     : xiaohao Z
 * Blog     : http://www.cnblogs.com/shu-xiaohao/
 * Last modified : 2014-02-05 20:31
 * Filename     : uva_10600.cpp
 * Description     : 
 * ************************************************/

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#define MP(a, b) make_pair(a, b)
#define PB(a) push_back(a)

using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<unsigned int,unsigned int> puu;
typedef pair<int, double> pid;
typedef pair<ll, int> pli;
typedef pair<int, ll> pil;

const int INF = 0x3f3f3f3f;
const double eps = 1E-6;
const int LEN = 1001;
struct edge{
     int fr, to, val;
}e[LEN*LEN];
int n, m, parent[LEN], vis[LEN*LEN];

bool cmp(edge a, edge b){return a.val < b.val;}
void init(){for(int i=0; i<LEN; i++) parent[i] = i;}
int Find(int x){return parent[x] == x? x: parent[x] = Find(parent[x]);}

int kruskal(int pos){
     int cnt = 0, ret = 0;
     for(int i=0; i<m; i++){
        if(i == pos) continue;
        int pa = Find(e[i].fr), pb = Find(e[i].to);
        if(pa == pb) continue;
        parent[pa] = pb;
        if(pos < 0)vis[i] = 1;
        ret += e[i].val;
        cnt ++;
        if(cnt == n-1) break;
     }
     if(cnt!=n-1) return INF;
     return ret;
}

int main()
{
//    freopen("in.txt", "r", stdin);

    int a, b, v, T;
    scanf("%d", &T);
    while(T--){
        init();
        scanf("%d%d", &n, &m);
        for(int i=0; i<m; i++)
             scanf("%d%d%d", &e[i].fr, &e[i].to, &e[i].val);
        sort(e, e+m, cmp);
        memset(vis, 0, sizeof vis);
        printf("%d ", kruskal(-1));
        int ans = INF;
        for(int i=0; i<m; i++)
            if(vis[i]) {
                init();
                ans = min(ans, kruskal(i));
            }
        printf("%d
", ans);
    }
    return 0;
}