hdu 1102 prim算法

最新推荐文章于 2020-04-26 13:16:04 发布

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本文介绍了一道关于最小生成树的经典问题，并通过Prim算法求解最优路径总长度。题目要求连接所有村庄使得总距离最短，已知部分村庄间道路已建成。

Problem Description
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

题解：

prim算法模板题。注意已经造好的路，cost为0.
最小生成树，将已有的路径花费置为0。

代码：

#include <bits/stdc++.h>

using namespace std;
const int INF = 0x3f3f3f3f;
const int MAXN = 110;

bool vis[MAXN];
int lowc[MAXN];
int cost[MAXN][MAXN];

void update(int x,int y,int v)
{
    cost[x-1][y-1]=v;
    cost[y-1][x-1]=v;
    return ;
}

int Prim(int cost[][MAXN],int n)
{
    int ans=0;
    memset(vis,false,sizeof(vis));
    vis[0]=true;
    for(int i=1;i<n;i++)
    {
        lowc[i]=cost[0][i];
    }
    for(int i=1;i<n;i++)
    {
        int minc = INF;
        int p = -1;
        for(int j=0;j<n;j++)
        {
            if(!vis[j]&&minc>lowc[j])
            {
                minc = lowc[j];
                p = j;
            }
        }
        if(minc==INF)
        {
            return -1;
        }
        ans+=minc;
        vis[p]=true;
        for(int j=0;j<n;j++)
        {
            if(!vis[j]&&lowc[j]>cost[p][j])
            {
                lowc[j]=cost[p][j];
            }
        }
    }
    return ans;
}

int main()
{
    int N,Q;
    while(cin>>N)
    {
        memset(cost,0x3f,sizeof(cost));
        for(int i=0;i<N;i++)
            for(int j=0;j<N;j++)
        {
            scanf("%d",&cost[i][j]);
        }

        cin>>Q;
        int u,v;
        for(int i=0;i<Q;i++)
        {
            scanf("%d%d",&u,&v);
            update(u,v,0);
        }
        cout<<Prim(cost,N)<<endl;
    }
    return 0;
}