暑假集训每日一题0723 （最小生成树）

Description

Farmer John has been elected mayor of his town! One of his campaign promises was to bring internet connectivity to all farms in the area. He needs your help, of course.

Farmer John ordered a high speed connection for his farm and is going to share his connectivity with the other farmers. To minimize cost, he wants to lay the minimum amount of optical fiber to connect his farm to all the other farms.

Given a list of how much fiber it takes to connect each pair of farms, you must find the minimum amount of fiber needed to connect them all together. Each farm must connect to some other farm such that a packet can flow from any one farm to any other farm.

The distance between any two farms will not exceed 100,000.

Input

There is several test cases.
Line 1:     The number of farms, N (3 <= N <= 100).
Line 2..end:     The subsequent lines contain the N x N connectivity matrix, where each element shows the distance from on farm to another. Logically, they are N lines of N space-separated integers. Physically, they are limited in length to 80 characters, so some lines continue onto others. Of course, the diagonal will be 0, since the distance from farm i to itself is not interesting for this problem.

Output

The single output contains the integer length that is the sum of the minimum length of fiber required to connect the entire set of farms.

Sample Input

4

0 4 9 21

4 0 8 17

9 8 0 16

21 17 16 0

Sample Output

28

裸的最小生成树。

View Code

#include <stdio.h>

#include <stdlib.h>

#define N 110

#define M 10100

int n;

int d[N][N];

int p[N];

struct node

{

    int i,j,d;

};

node edge[M];

int cnt;

void make_set()

{

    for(int i=0;i<n;i++)    p[i]=i;

}

int find_set(int i)

{

    if(p[i]!=i) p[i]=find_set(p[i]);

    return p[i];

}

void union_set(int i,int j)

{

    i=find_set(i);

    j=find_set(j);

    p[j]=i;

}

int cmp(const void *a,const void *b)

{

    node *x=(node *)a;

    node *y=(node *)b;

    return (x->d)>(y->d)?1:-1;

}

int main()

{

    int i,j,k,ans;

    while(~scanf("%d",&n))

    {

        for(i=0;i<n;i++)

        {

            for(j=0;j<n;j++)    scanf("%d",&d[i][j]);

        }

        cnt=0;

        for(i=0;i<n;i++)

        {

            for(j=0;j<n;j++)

            {

                edge[cnt].i=i;

                edge[cnt].j=j;

                edge[cnt++].d=d[i][j];

            }

        }

        qsort(edge,cnt,sizeof(edge[0]),cmp);

        make_set();

        ans=0;

        for(k=0;k<cnt;k++)

        {

            i=edge[k].i;

            j=edge[k].j;

            if(find_set(i)^find_set(j)) ans+=edge[k].d,union_set(i,j);

        }

        printf("%d\n",ans);

    }

    return 0;

}