uva 10600 最小与次小生成树

最新推荐文章于 2022-02-09 21:32:53 发布

sdjzujxc

最新推荐文章于 2022-02-09 21:32:53 发布

阅读量968

点赞数

CC 4.0 BY-SA版权

分类专栏：生成树

本文链接：https://blog.youkuaiyun.com/sdjzujxc/article/details/8776712

生成树专栏收录该内容

4 篇文章

订阅专栏

本文探讨了为确保学校电力供应稳定而设计的最经济连接方案，通过寻找最小生成树与次小生成树来解决市长面临的难题。利用图算法解决实际问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Problem A

ACM CONTEST AND BLACKOUT

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 A_i, B_i, C_i , where C_i is the cost of the connection (1£C_i£300) between schools A_i and B_i. 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

Sample Output

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

110 121

37 37

求最小生成树与次小生成树。

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define N 105
#define inf 99999999
int map[N][N],F[N][N];//F存储第i个节点到第j个节点的路径中的最大权值
int vis[N],dis[N],pre[N];
int use[N][N];//标记边Eij,0表示不存在，1表示用过，2表示存在没用过
int n,m;
int sum1,sum2;
int prime()
{
    sum1=0;
    memset(vis,0,sizeof(vis));
    int k=1;
    for(int i=1; i<=n; i++)
    {
        vis[i]=0;
        dis[i]=map[1][i];
        pre[i]=1;
    }
    pre[1]=-1;
    vis[1]=1;
    for(int i=1; i<n; i++)
    {
        int mmin=inf;
        for(int j=1; j<=n; j++)
        {
            if(!vis[j]&&dis[j]<mmin)
            {
                mmin=dis[j];
                k=j;
            }
        }
        if(pre[k]!=-1)
        {
            use[pre[k]][k]=use[k][pre[k]]=1;
            for(int j=1; j<=n; j++)
            {
                if(vis[j])
                F[j][k]=max(F[j][pre[k]],map[pre[k]][k]);
            }
        }
        vis[k]=1;
        sum1+=mmin;
        for(int j=1;j<=n;j++)
        {
            if(!vis[j]&&dis[j]>map[k][j])
            {
                dis[j]=map[k][j];
                pre[j]=k;
            }
        }
    }
    return sum1;
}
int second()
{
    sum2=inf;
    for(int i=1;i<=n;i++)
    for(int j=1;j<=n;j++)
    {
        if(use[i][j]==2&&sum1+map[i][j]-F[i][j]<sum2)
        {
            sum2=sum1+map[i][j]-F[i][j];
        }
    }
    return sum2;
}

int main()
{
    int cas,a,b,v;
    cin>>cas;
    while(cas--)
    {
        cin>>n>>m;
        for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
        {
            map[i][j]=inf;
            F[i][j]=0;
            use[i][j]=0;
        }
        for(int i=0;i<m;i++)
        {
            scanf("%d%d%d",&a,&b,&v);
            map[a][b]=map[b][a]=v;
            use[a][b]=use[b][a]=2;
        }
        int s1=prime();
        int s2=second();
        printf("%d %d\n",s1,s2);
    }
    return 0;
}
/*
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
*/