HDU 3371 Connect the Cities MST存在已建好的边

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最新推荐文章于 2021-04-08 19:17:52 发布

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本文讨论了一个城市重建问题，目标是在给定数量的城市中，通过选择最少的边建立连接，以重新构建城市的网络，并且计算所需的最低成本。文章详细介绍了输入格式、数据限制以及输出要求，最终提供了求解此问题的一种算法实现。

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Connect the Cities

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7040 Accepted Submission(s): 2008

Problem Description

In 2100, since the sea level rise, most of the cities disappear. Though some survived cities are still connected with others, but most of them become disconnected. The government wants to build some roads to connect all of these cities again, but they don’t want to take too much money.

Input

The first line contains the number of test cases.
Each test case starts with three integers: n, m and k. n (3 <= n <=500) stands for the number of survived cities, m (0 <= m <= 25000) stands for the number of roads you can choose to connect the cities and k (0 <= k <= 100) stands for the number of still connected cities.
To make it easy, the cities are signed from 1 to n.
Then follow m lines, each contains three integers p, q and c (0 <= c <= 1000), means it takes c to connect p and q.
Then follow k lines, each line starts with an integer t (2 <= t <= n) stands for the number of this connected cities. Then t integers follow stands for the id of these cities.

Output

For each case, output the least money you need to take, if it’s impossible, just output -1.

Sample Input

Sample Output

Author

dandelion

Source

HDOJ Monthly Contest – 2010.04.04

Recommend

lcy

题意是说有n个城市，其中有m条边未建，然后还有已经建好的。让你求最少费用。

一开始用prime做超时，以为次算法不行，便用Kruskal做了一遍也是超时。最后才知道原来要用C++交，而不是G++。

晕哒哒~~

#include<stdio.h>
#include<string.h>
#define inf 1000
#define M 507
bool vis[M];
int dis[M],map[M][M];
int n,m,k;

void prime()
{
    int minn,id,count=0;
    memset(vis,false,sizeof(vis));
    for(int i=1; i<=n; i++)
        dis[i]=map[1][i];
    vis[1]=true;
    for(int i=2; i<=n; i++)
    {
        minn=inf;
        id=0;
        for(int j=2; j<=n; j++)
            if(!vis[j]&&minn>dis[j])
            {
                minn=dis[j];
                id=j;
            }
            if(minn==inf)break;
            vis[id]=true;
            
            count+=minn;
            for(int j=2; j<=n; j++)
                if(!vis[j]&&dis[j]>map[j][id])
                    dis[j]=map[j][id];
    }
    if(minn!=inf)
        printf("%d\n",count);
    else
        printf("-1\n");
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d%d",&n,&m,&k);
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=n; j++)
                map[i][j]=inf;
            map[i][i]=0;
        }
        int u,v,c,num,city,city1;
        for(int i=1; i<=m; i++)
        {
            scanf("%d%d%d",&u,&v,&c);
            map[v][u]=map[v][u]>c?c:map[v][u];
            map[u][v]=map[v][u];
        }
        for(int i=1; i<=k; i++)
        {
            scanf("%d %d",&num,&city1);
            for(int j=1; j<num; j++)
            {
                scanf("%d",&city);
                map[city1][city]=map[city][city1]=0;
                city1=city;
            }
        }
        prime();
    }
    return 0;
}