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 题解：让我们设想一下这样的一个无向图：在征募某个人a时，如果使用了a和b之间的关系，那么就连一条a到b的边。假设这个图存在圈，那么无论以什么样的顺序征募这个圈上所有的人，都会产生矛盾，因此可以知道这个图是一片森林。反之，如果给了一片森林就可以使用对应的关系确定征募的顺序。因此把人看作顶点，关系看作边，那么这个问题就可以转化为求解无向图中的最大权森林问题，也就是把所有边权取反之后用最小生成树算法求解。--------------------------------------摘自《挑战程序设计竞赛》
1 #include<cstdio>
2 #include<cstring>
3 #include<iostream>
4 #include<algorithm>
5 using namespace std;
6 typedef long long ll;
7
8 const int maxn=50005;
9
10 struct edge{
11 int u,v,cost;
12 bool operator<(const edge& i)const{
13 return cost<i.cost;
14 }
15 };
16
17 edge es[maxn];
18 int n,m,R,F[maxn];
19
20 int Find(int a){
21 if(a!=F[a]) F[a]=Find(F[a]);
22 return F[a];
23 }
24
25 bool unite(int a,int b){
26 int x=Find(a),y=Find(b);
27 if(x!=y) { F[x]=y; return true; }
28 else return false;
29 }
30
31 int Kruskal(){
32 sort(es,es+R);
33 int ans=0;
34 for(int i=0;i<R;i++){
35 edge e=es[i];
36 if(unite(e.u,e.v)) ans=ans+e.cost;
37 }
38 return ans;
39 }
40
41 int main()
42 { int kase;
43 cin>>kase;
44 while(kase--){
45 cin>>n>>m>>R;
46 int x,y,most;
47 for(int i=0;i<R;i++){
48 scanf("%d%d%d",&x,&y,&most);
49 es[i]=(edge){x,y+n,-most};
50 }
51 for(int i=0;i<n+m;i++) F[i]=i;
52 int ans=10000*(n+m)+Kruskal();
53 cout<<ans<<endl;
54 }
55 return 0;
56 }