POJ - 3723 Conscription POJ（最小生成树）

最新推荐文章于 2019-08-03 17:09:44 发布

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本文通过一个具体的案例，介绍如何使用最小生成树算法解决实际问题。主人公Windy希望组建一支军队保护国家，需要从选拔出的人员中挑选士兵，并利用他们之间的特定关系减少成本。文章详细解释了算法的具体实现过程，包括数据结构定义、关键函数实现及输入输出样例。

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 x_i, y_i and d_i.
There is a blank line before each test case.

1 ≤ N, M ≤ 10000
0 ≤ R ≤ 50,000
0 ≤ x_i < N
0 ≤ y_i < M
0 < d_i < 10000

Output

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

Sample Input

Sample Output

71071
54223

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<iostream>
#include<math.h>
#include<queue>
#include<map>
#include<vector>
using namespace std;
#define inf 0x3f3f3f3f
#define ll long long
int n,m,r,st,tot;
struct node
{
    int u,v,w;
}e[6000000];
int fa[6000000];
void add(int u,int v,int w)
{
    e[tot].u=u;
    e[tot].v=v;
    e[tot++].w=w;
}
bool cmp(node n1,node n2)
{
    return n1.w<n2.w;
}
int find(int x)
{
    if(fa[x]==x) return x;
    else return fa[x]=find(fa[x]);
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        tot=0;
        scanf("%d%d%d",&n,&m,&r);
        st=n+m;
        int u,v,w;
        for(int i=0;i<=n+m;i++)
        {
          add(st,i,10000);//额外加1个点 到所有的点的距离都为10000
          fa[i]=i;
        }
        for(int i=1;i<=r;i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            add(u,v+n,10000-w);
        }
        sort(e,e+tot,cmp);
        int cnt=0;
        ll ans=0;
        for(int i=0;i<tot;i++)
        {
             u=e[i].u;
             v=e[i].v;
             w=e[i].w;
             int t1=find(u);
             int t2=find(v);
             if(t1!=t2)
             {
                 ans+=w;
                 fa[t1]=t2;
                 cnt++;
             }
             if(cnt==n+m) break;
        }
        printf("%lld\n",ans);

    }
}