最小生成树 - D - Constructing Roads

最新推荐文章于 2020-02-24 14:33:57 发布

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本文介绍了一道关于使用Prim算法求解最小生成树的问题。题目要求在已知部分村庄间道路的基础上，通过构建额外的道路使得所有村庄相连，并使新增道路的总长度最短。文章提供了详细的解题思路及AC代码。

D - Constructing Roads

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：给定N个点与各边的权值，已知一定的边已连接，求连接剩下的点的最小权值和。

题解：将已联系的边的权值改为0即可，prim算法。

AC代码：

#include<iostream>
using namespace std;
const int inf = 1e9;
int a[105][105];
int n,m,sum;
int vis[105];
int length[111];
void prim()
{
    for(int i=1;i<=n;i++)
    {
        length[i]=a[1][i];
        vis[i]=0;
    }
    for(int i=1;i<=n;i++)
    {
        int u=inf;
        int v;
        for(int j=1;j<=n;j++)
        {
            if(vis[j]==0&&u>length[j])
            {
                v=j;
                u=length[j];
            }
        }
        vis[v]=1;
        sum+=length[v];
        for(int j=1;j<=n;j++)
        {
            if(vis[j]==0&&length[j]>a[v][j])
            {
                length[j]=a[v][j];
            }
        }
    }
}
int main()
{
    while(cin>>n)
    {
        sum=0;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
        {
             cin>>a[i][j];
        }
        cin>>m;
        for(int i=1;i<=m;i++)
        {
           int x1,x2;
           cin>>x1>>x2;
           a[x1][x2]=a[x2][x1]=0;
        }
        prim();
        cout<<sum<<endl;
    }
    return 0;
}