poj-3723-Conscription

喜欢ヅ旅行 2022-05-27 04:48 124阅读 0赞

Conscription

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;65536K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>15768</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;5453</td> 
  </tr> 
 </tbody> 
</table>

Description

Windy has a country, and he wants to build an army to protect his country. He has picked up *N* girls and *M* boys and wants to collect them to be his soldiers. To collect a soldier without any privilege, he must pay 10000 RMB. There are some relationships between girls and boys and Windy can use these relationships to reduce his cost. If girl **x** and boy **y** have a relationship *d* and one of them has been collected, Windy can collect the other one with 10000-*d* RMB. Now given all the relationships between girls and boys, your assignment is to find the least amount of money Windy has to pay. Notice that only one relationship can be used when collecting one soldier.

Input

The first line of input is the number of test case.  
The first line of each test case contains three integers, *N*, *M* and *R*.  
Then *R* lines followed, each contains three integers *xi*, *yi* and *di*.  
There is a blank line before each test case.

1 ≤ *N*, *M* ≤ 10000  
0 ≤ *R* ≤ 50,000  
0 ≤ *xi* < *N*  
0 ≤ *yi* < *M*  
0 < *di* < 10000

Output

For each test case output the answer in a single line.

Sample Input

2
    
    5 5 8
    4 3 6831
    1 3 4583
    0 0 6592
    0 1 3063
    3 3 4975
    1 3 2049
    4 2 2104
    2 2 781
    
    5 5 10
    2 4 9820
    3 2 6236
    3 1 8864
    2 4 8326
    2 0 5156
    2 0 1463
    4 1 2439
    0 4 4373
    3 4 8889
    2 4 3133

Sample Output

71071
    54223

题目大意：一个国王要招兵，每个兵花费10000，国王发现若a和b之间有关系则可以利用a,b之间的关系使自己减少费用，但是每个关系只能使用一次，求国王花了最少的钱

代码：

#include<queue>
    #include<cmath>
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    int fu[10010*4];
    struct ss
    {
        int u,v,w;
    }aa[10010*8];
    int cmp(ss A,ss B)
    {
        return A.w>B.w;
    }
    int findroot(int x)
    {
        if(fu[x]!=x)
            fu[x]=findroot(fu[x]);
        return fu[x];
    }
    int hebing(int a,int b)
    {
        int x,y;
        x=findroot(a);
        y=findroot(b);
        if(x==y)
            return 0;
        fu[y]=x;
        return 1;
    }
    int main()
    {
        int test;
        scanf("%d",&test);
        while(test--)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            int i,j,ans=0,t1,t2,t3;
            for(i=0;i<=a+b+1;i++)
                fu[i]=i;
            for(i=0;i<c;i++)
            {
                scanf("%d%d%d",&t1,&t2,&t3);
                aa[i].w=t3;
                aa[i].u=t1;
                aa[i].v=t2+a;
            }
            sort(aa,aa+c,cmp);
            for(i=0;i<c;i++)
            {
                if(hebing(aa[i].u,aa[i].v))
                    ans+=aa[i].w;
            }
            printf("%d\n",(a+b)*10000-ans);
        }
    }