Time Limit: 1000MS Memory Limit: 65536K
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
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
运用kruskal算法的最大生成树的基础题，直接1A。
具体请参考：http://www.cnblogs.com/liangrx06/p/5083763.html
1 #include<cstdio> 2 #include<algorithm> 3 using namespace std; 4 int n,m,r; 5 struct Node{ 6 int x,y,d; 7 }node[50000+5]; 8 int par[20000+5]; 9 bool cmp(Node x,Node y){return x.d>y.d;} 10 void init() 11 { 12 sort(node+1,node+r+1,cmp);//按d从大到小排列，以满足最大生成树的要求 13 for(int i=1;i<=n+m;i++) par[i]=i;//初始化并查集 14 } 15 int find(int x){return( par[x]==x ? x : par[x]=find(par[x]) );} 16 int kruskal() 17 { 18 int d_sum=0; 19 for(int i=1;i<=r;i++) 20 { 21 int x=find(node[i].x),y=find(node[i].y); 22 if(x != y)//找到一条属于最大生成树的新边 23 { 24 par[y]=x;//合并 25 d_sum+=node[i].d;//这条边属于最大生成树，即对应的题目中要使用的“关系”( ‵▽′)ψ 26 } 27 } 28 return d_sum; 29 } 30 int main() 31 { 32 int t; 33 scanf("%d",&t); 34 while(t--) 35 { 36 scanf(" %d%d%d",&n,&m,&r);// n个girl , m个boy , r句话 37 for(int i=1;i<=r;i++) 38 { 39 int x,y,d; 40 scanf(" %d%d%d",&x,&y,&d); //girl - x ; boy - y 41 node[i].x=x+1,node[i].y=n+1+y;//这样可以确保男女孩的编号必然是不同的 42 node[i].d=d; 43 } 44 init(); 45 int d_sum=kruskal(); 46 printf("%d ",(n+m)*10000 - d_sum); 47 } 48 }