Constructing Roads

最新推荐文章于 2024-07-21 14:30:00 发布

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本文介绍了一种通过使用Prime算法和Kruskal算法解决最小生成树问题的方法，旨在找到连接所有节点且总边权最小的树形结构。适用于给定村庄间距离及已建成道路的情况下，如何构建新的道路以确保所有村庄连通且总建设成本最低。

There are N villages, which are numbered from 1 to N, and you should build some roads such that every two villages can connect to each other. We say two village A and B are connected, if and only if there is a road between A and B, or there exists a village C such that there is a road between A and C, and C and B are connected.

We know that there are already some roads between some villages and your job is the build some roads such that all the villages are connect and the length of all the roads built is minimum.

Input

The first line is an integer N (3 <= N <= 100), which is the number of villages. Then come N lines, the i-th of which contains N integers, and the j-th of these N integers is the distance (the distance should be an integer within [1, 1000]) between village i and village j.

Then there is an integer Q (0 <= Q <= N * (N + 1) / 2). Then come Q lines, each line contains two integers a and b (1 <= a < b <= N), which means the road between village a and village b has been built.

Output

You should output a line contains an integer, which is the length of all the roads to be built such that all the villages are connected, and this value is minimum.

Sample Input

Sample Output

题意：有n个城市，城市标号1~n，已知每两个城市之间连接所需的费用以及已经连通的城市，要使所有的城市连通所需的费用最小。

代码①号（prime）：

#include<stdio.h>
#define inf 0x3f3f3f
int n,m,x,y,a[120][120];
void prime()
{
    int i,j,k,mi,cnt=0,sum=0,dis[120],vis[120]={0};
    for(int i=1;i<=n;i++)//dis[i]表示连接城市1到i城市的所需的费用
        dis[i]=a[1][i];
    vis[1]=1;
    cnt++;
    while(cnt<n)//有n个城市，所以总共有n-1次连接就好
    {
        mi=inf;
        for(i=1;i<=n;i++)
        {
            if(vis[i]==0&&dis[i]<mi)
            {
                mi=dis[i];
                j=i;
            }
        }
        vis[j]=1;
        cnt++;
        sum+=dis[j];
        for(k=1;k<=n;k++)
        {
            if(vis[k]==0&&dis[k]>a[j][k])
                dis[k]=a[j][k];
        }
    }
    printf("%d\n",sum);
}
int main()
{
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
                scanf("%d",&a[i][j]);
        }
        scanf("%d",&m);
        for(int i=0;i<m;i++)
        {
            scanf("%d %d",&x,&y);
            a[x][y]=a[y][x]=0;
        }
        prime();
    }
    return 0;
}

代码②号（kruskal）：

#include<stdio.h>
#include<algorithm>
using namespace std;
int n,m,t,x,y,f[12000];
struct node
{
    int x,y,d;
}p[12000];
bool cmp(node a,node b)
{
    return a.d<b.d;
}
int fond(int x)
{
    while(x!=f[x])
        x=f[x];
    return x;
}
void bing(int a,int b)
{
    a=fond(a);
    b=fond(b);
    if(a!=b)
        f[a]=b;
}
void kruskal()
{
    int sum=0,num=0;
    for(int i=0;i<t;i++)
    {
        if(fond(p[i].x)!=fond(p[i].y))
        {
            bing(p[i].x,p[i].y);
            sum+=p[i].d;
            num++;
        }
    }
    printf("%d\n",sum);
}
int main()
{
    while(~scanf("%d",&n))
    {
        t=0;
        for(int i=1;i<=n;i++)
        {
            f[i]=i;
            for(int j=1;j<=n;j++)
            {
                scanf("%d",&p[t].d);
                p[t].x=i;
                p[t].y=j;
                t++;
            }
        }
        scanf("%d",&m);
        for(int i=0;i<m;i++)
        {
            scanf("%d%d",&x,&y);//因为城市x到城市y已经连接，所以将它们并到一起
            bing(x,y);
        }
        sort(p,p+t,cmp);
        kruskal();
    }
    return 0;
}