HDU - 1102 - Constructing Roads

最新推荐文章于 2022-08-16 15:04:01 发布

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最新推荐文章于 2022-08-16 15:04:01 发布

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分类专栏：最小生成树

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本文介绍了一个经典的最小生成树问题，给定若干村庄及其之间的距离，部分村庄已通过道路相连，任务是构造剩余的道路使得所有村庄连通且总距离最短。文章提供了详细的解题思路及C++实现代码。

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Constructing Roads HDU - 1102

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
3
0 990 692
990 0 179
692 179 0
1
1 2
Sample Output
179

题意：给一定数目的村庄与他们的距离，有些村庄已经连接起来，求最短连接距离。

题解：最小生成树模板题，注意连接起来的村庄是无向的即可。

AC代码：

#include<cstdio>
#include<cmath>
#include<iostream>
using namespace std;
const int maxn = 105;
int n,m;
int a[maxn][maxn];
int vis[maxn];
int length[maxn];
int sum;
void prim()
{
    for(int i=1;i<=n;i++)
    {
        length[i]=a[1][i];
        vis[i]=0;
    }
    vis[1]=1;
    for(int i=1;i<=n;i++)
    {
        int u=1e9;
        int v;
        for(int j=1;j<=n;j++)
        {
            if(vis[j]==0&&u>length[j])
            {
                u=length[j];
                v=j;
            }
        }
        vis[v]=1;
        sum+=length[v];
        for(int j=1;j<=n;j++)length[j]=min(length[j],a[v][j]);
    }
}
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        sum=0;
        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=1;i<=m;i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            a[x][y]=a[y][x]=0;
        }
        prim();
        printf("%d\n",sum);
    }
    return 0;
}