POJ 3723 Conscription(最大生成ツリー)

Conscription
Time Limit: 1000MS
 
Memory Limit: 65536K
Total Submissions: 5995
 
Accepted: 2058
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 ≤ 100000 ≤ R ≤ 50,0000 ≤ xi < N0 ≤ yi < M0 < 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

Source
POJ Monthly Contest – 2009.04.05 , windy7926778
 
 
Windyは軍隊を設立するには、N人の女子学生とM人の男子学生を募集して構成し、一人当たり10000 RMBを払う必要がある.現在、彼らの中にはR組の関係、すなわちx番目の女子学生とy番目の男子学生の関係がdであることが知られており、先にそのうちの1つを募集し、彼らの関係を利用してもう1つを募集すれば、安いRMBを募集することができ、Windyに少なくともどのくらいのRMBでこの軍隊を作ることができるかを聞く.
 
考え方:基礎的なkruskal.ここでは図が必ずしもつながっているとは限らないので、最大生成森林を求めます.
 

#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>



using namespace std;



const int N=10010;

const int M=50010;



struct Edge{

    int u,v;

    int cap;

}edge[M<<2];



int n,m,ans,cnt;

int father[M];



void addedge(int cu,int cv,int cw){

    edge[cnt].u=cu;     edge[cnt].v=cv;     edge[cnt].cap=cw;

    cnt++;

}



void makeSet(){

    for(int i=1;i<M;i++){

        father[i]=i;

    }

}



int findSet(int x){

    if(x!=father[x]){

        father[x]=findSet(father[x]);

    }

    return father[x];

}



int cmp(Edge a,Edge b){

    return a.cap>b.cap;     //// ans

}



void Kruskal(){

    sort(edge,edge+cnt,cmp);

    for(int i=0;i<cnt;i++){

        int fx=findSet(edge[i].u);

        int fy=findSet(edge[i].v);

        if(fx!=fy){     // ， 

            father[fy]=fx;

            ans-=edge[i].cap;

        }

    }

}



int main(){



    freopen("input.txt","r",stdin);



    int t;

    scanf("%d",&t);

    while(t--){

        cnt=0;

        makeSet();

        int q;

        scanf("%d%d%d",&n,&m,&q);

        int u,v,w;

        while(q--){

            scanf("%d%d%d",&u,&v,&w);

            addedge(u,v+n,w);

        }

        ans=(n+m)*10000;

        Kruskal();

        printf("%d
",ans);

    }

    return 0;

}