UVA10600 ACM Contest and Blackout —— 次小生成树

题目链接：https://vjudge.net/problem/UVA-10600

In order to prepare the “The First National ACM School Contest” (in 20??) the major of the city decided to provide all the schools with a reliable source of power. (The major is really afraid of blackoutsJ). So, in order to do that, power station “Future” and one school (doesn’t matter which one) must be connected; in addition, some schools must be connected as well. You may assume that a school has a reliable source of power if it’s connected directly to “Future”, or to any other school that has a reliable source of power. You are given the cost of connection between some schools. The major has decided to pick out two the cheapest connection plans – the cost of the connection is equal to the sum of the connections between the schools. Your task is to help the major — find the cost of the two cheapest connection plans.

Input

The Input starts with the number of test cases, T (1 < T < 15) on a line. Then T test cases follow. The first line of every test case contains two numbers, which are separated by a space, N (3 < N < 100) the number of schools in the city, and M the number of possible connections among them. Next M lines contain three numbers Ai , Bi , Ci , where Ci is the cost of the connection (1 < Ci < 300) between schools Ai and Bi . The schools are numbered with integers in the range 1 to N.

Output

For every test case print only one line of output. This line should contain two numbers separated by a single space – the cost of two the cheapest connection plans. Let S1 be the cheapest cost and S2 the next cheapest cost. It’s important, that S1 = S2 if and only if there are two cheapest plans, otherwise S1 < S2. You can assume that it is always possible to find the costs S1 and S2. Sample Input 2 5 8 1 3 75 3 4 51 2 4 19 3 2 95 2 5 42 5 4 31 1 2 9 3 5 66 9 14 1 2 4 1 8 8 2 8 11 3 2 8 8 9 7 8 7 1 7 9 6 9 3 2 3 4 7 3 6 4 7 6 2 4 6 14 4 5 9 5 6 10 Sample Output 110 121 37 37

题解：

赤裸裸的求最小生成树和次小生成树。

代码如下：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long LL;
const double EPS = 1e-6;
const int INF = 2e9;
const LL LNF = 9e18;
const int MOD = 1e9+7;
const int MAXN = 1e2+10;

int cost[MAXN][MAXN], lowc[MAXN], pre[MAXN], Max[MAXN][MAXN];
bool  vis[MAXN], used[MAXN][MAXN];

int Prim(int st, int n)
{
    int ret = 0;
    memset(vis, false, sizeof(vis));
    memset(used, false, sizeof(used));
    memset(Max, 0, sizeof(Max));

    for(int i = 1; i<=n; i++)
        lowc[i] = (i==st)?0:INF;
    pre[st] = st;

    for(int i = 1; i<=n; i++)
    {
        int k, minn = INF;
        for(int j = 1; j<=n; j++)
            if(!vis[j] && minn>lowc[j])
                minn = lowc[k=j];

        vis[k] = true;
        ret += minn;
        used[pre[k]][k] = used[k][pre[k]] = true;  //pre[k]-k的边加入生成树
        for(int j = 1; j<=n; j++)
        {
            if(vis[j] && j!=k)  //如果遇到已经加入生成树的点，则找到两点间路径上的最大权值。
                Max[j][k] = Max[k][j] = max(Max[j][pre[k]], lowc[k]);   //k的上一个点是pre[k]
            if(!vis[j] && lowc[j]>cost[k][j]) //否则，进行松弛操作
            {
                lowc[j] = cost[k][j];
                pre[j] = k;
            }
        }
    }
    return (ret==INF)?-1:ret;
}

int SMST(int t1 ,int n)
{
    int ret = INF;
    for(int i = 1; i<=n; i++)   //用生成树之外的一条边去代替生成树内的一条边
    for(int j = i+1; j<=n; j++)
    {
        if(cost[i][j]!=INF && !used[i][j])  //去掉了i-j路径上的某条边，但又把i、j直接连上，所以还是一棵生成树。
            ret = min(ret, t1+cost[i][j]-Max[i][j]);
    }
    return ret;
}

int main()
{
    int T, n, m;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i = 1; i<=n; i++)
        for(int j = 1; j<=n; j++)
            cost[i][j] = (i==j)?0:INF;

        for(int i = 1; i<=m; i++)
        {
            int u, v, w;
            scanf("%d%d%d", &u, &v, &w);
            cost[u][v] = cost[v][u] = w;
        }

        int t1 = Prim(1, n);
        int t2 = SMST(t1, n);
        printf("%d %d
", t1, t2);
    }
}