【杭电2015年12月校赛D】【水题 最小生成树】Happy Value 最小生成树裸题

Happy Value

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1388    Accepted Submission(s): 408

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

20 11 030 1 51 0 35 3 0

 

Sample Output

18

 

#include<stdio.h>#include<algorithm>#include<ctype.h>#include<string.h>using namespace std;int casenum,casei;typedef long long LL;const int N=105;const int M=105*105;int n,m;struct A{    int x,y,z;    bool operator < (const A& b)const    {        return z>b.z;    }}a[M];int f[N];int find(int x){return f[x]==x?x:f[x]=find(f[x]);}int main(){    while(~scanf("%d",&n))    {        m=0;        for(int i=1;i<=n;++i)        {            f[i]=i;            for(int j=1;j<=n;++j)            {                int x;                scanf("%d",&x);                if(i<j)                {                    ++m;                    a[m].x=i;                    a[m].y=j;                    a[m].z=x;                }            }        }        int ans=0;        sort(a+1,a+m+1);        for(int i=1;i<=m;++i)        {            int x=find(a[i].x);            int y=find(a[i].y);            if(x!=y)            {                ans+=a[i].z;                f[y]=x;            }        }        printf("%d\n",ans);    }    return 0;}/*【题意】就是一个完全图的最小生成树。什么也不说了= = */