Agri-Net（最小生成树）

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第一道图论题目，挺简单的但是做了好久，现在才知道即使是因为没有while（~scanf（“%d”，&n））也会报错，而有些提目并不会把他体现出来

C - Agri-Net

Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%lld & %llu

Submit Status

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

The input includes several cases. For each case, the first line contains the number of farms, N (3 <= N <= 100). The following lines contain the N x N conectivity 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

For each case, output a single integer length that is the sum of the minimum length of fiber required to connect the entire set of farms.

Sample Input

40 4 9 214 0 8 179 8 0 1621 17 16 0

Sample Output

123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int e[110][110],dis[110],book[110];int sum;int inf=999999999;//定义无穷void prim(int n){    int count=0;    int i,j,k;
    //初始化dis数组，这里是一号顶点到各个点的初始距离，因为当前生成树就只有1号节点    for(i=1;i<=n;i++)    {        dis[i]=e[1][i];    }
    //核心代码
    //将1号顶点加入生成树    book[1]=1;//标记已经加入树    count++;    while(count<n)    {        int min=inf;        j=1;        for(i=1;i<=n;i++)        {            if(book[i]==0&&dis[i]<min)            {                min=dis[i];                j=i;            }        }        book[j]=1;count++;sum=sum+dis[j];
        //扫描当前顶点j的所有的边。再以j为中间点，更新生成树到每一个顶点的距离        for(k=1;k<=n;k++)        {            if(book[k]==0&&dis[k]>e[j][k])                dis[k]=e[j][k];        }    }}int main(){    int n,i,j;    while(~scanf("%d",&n))    {    sum=0;    memset(book,0,sizeof(book));
    //可加一个初始化    for(i=1;i<=n;i++)    {        for(j=1;j<=n;j++)        {            scanf("%d",&e[i][j]);
            //无向图的话需要反复存储，关于对角线对称 
            //例如a[i][j]=k;a[j][i]=k;根据输入而定        }    }    prim(n);    printf("%d\n",sum);    }    return 0;}

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