HDU计算机学院大学生程序设计竞赛(2015’12)Happy Value
Problem Description
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”.
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 Hij(0<=Hij<=10000) in ith row and jth column indicates that ith resident have a “Happy Value” Hij with jthresident. And Hij(i!=j) is equal to Hji. Hij(i=j) is always 0.
#include<stdio.h>
//#include<bits/stdc++.h>
#include<string.h>
#include<iostream>
#include<math.h>
#include<sstream>
#include<set>
#include<queue>
//#include<map>
#include<vector>
#include<algorithm>
#include<limits.h>
#define inf 0x3fffffff
#define INF 0x3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define ULL unsigned long long
using namespace std;
int map[110][110];
int dis[110];
int flag[110];
int n;
int Prime()
{
int i,j;
int pos,sum=0;
memset(flag,0,sizeof(flag));
for(i=1; i<=n; i++)
{
dis[i]=map[1][i];
}
flag[1]=1;
for(i=1; i<n; i++)
{
int mid=inf;
for(j=1; j<=n; j++)
{
if(!flag[j]&&dis[j]<mid)
{
mid=dis[j];
pos=j;
}
}
flag[pos]=1;
sum+=mid;
for(j=1; j<=n; j++)
{
if(!flag[j]&&map[pos][j]<dis[j])
{
dis[j]=map[pos][j];
}
}
}
return sum;
}
int main()
{
int i,j;
int q,a,b;
while(~scanf("%d",&n))
{
memset(map,0,sizeof(map));
int ans;
for(i=1; i<=n; i++)
{
for(j=1; j<=n; j++)
{
scanf("%d",&ans);
map[i][j]=-ans;
}
}
printf("%d\n",-Prime());
}
return 0;
}
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