電腦學院大學生程式設計競賽（2015’12）Happy Value

Happy Value Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1337    Accepted Submission(s): 392


Problem Description In an apartment, there are N residents. The Internet Service Provider (ISP) wants to connect these residents with N – 1 cables. 
However, the friendships of the residents are different. There is a “Happy Value” indicating the degrees of a pair of residents. The higher “Happy Value” is, the friendlier a pair of residents is. So the ISP wants to choose a connecting plan to make the highest sum of “Happy Values”.  
Input There are multiple test cases. Please process to end of file.
For each case, the first line contains only one integer N (2<=N<=100), indicating the number of the residents.
Then N lines follow. Each line contains N integers. Each integer H ij(0<=H ij<=10000) in i th row and j th column indicates that i th resident have a “Happy Value” H ijwith j th resident. And H ij(i!=j) is equal to H ji. H ij(i=j) is always 0.  
Output For each case, please output the answer in one line.  
Sample Input

2 0 1 1 0 3 0 1 5 1 0 3 5 3 0  
Sample Output

1 8  

總是望著曾經的空間發獃，那些說好不分開的朋友不在了，轉身，陌路。 熟悉的，安靜了， 安靜的，離開了， 離開的，陌生了， 陌生的，消失了， 消失的，陌路了。

#include <stdio.h>#include <string.h>#define MaxInt 9999999#define N 110int map[N][N],low[N],visited[N],n;int prim(){    int i,j,pos=1,min,result=0;    memset(visited,0,sizeof(visited));    visited[1]=1;    for(i=2; i<=n; i++)       low[i]=map[pos][i];    for(i=1; i<n; i++)    {        min=MaxInt;        for(j=1; j<=n; j++)            if(visited[j]==0&&min>low[j])            {                min=low[j];                pos=j;            }        result+=min;        visited[pos]=1;        for(j=1; j<=n; j++)            if(visited[j]==0&&low[j]>map[pos][j])                low[j]=map[pos][j];    }    return result;}int main(){    int i,v,j,ans;    while(scanf("%d",&n)!=EOF)    {        memset(map,MaxInt,sizeof(map));        for(i=1; i<=n; i++)            for(j=1; j<=n; j++)            {                scanf("%d",&v);                map[i][j]=-v;            }        ans=-1*prim();        printf("%d\n",ans);    }    return 0;}

@執念  "@☆但求“”安★ 下次我們做的一定會更好。。。。
為什 麼這次的題目是英文的。。。。QAQ...