Conscription(POJ 3723)

• 原题如下：
  Conscription

  Time Limit: 1000MS		Memory Limit: 65536K
  Total Submissions: 16584		Accepted: 5764

  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 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

  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：设想这样一个无向图：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，假设这个图中存在圈，那么无论以什么顺序征募这个圈上的所有人，都会产生矛盾，即由于圈的存在，矛盾是必然的，由此可以知道，这个图应该是一片森林。反过来，给定一片森林里，必然可以使用对应的关系确定征募的顺序。综上，把人看作点，关系看作边，这个问题就可以转化为求解无向图中的最大权森林问题，最大权森林问题可以通过把所有边权取反后用最小生成树的算法求解
• 代码：

   #include <cstdio>
   #include <queue>
   #include <vector>
   #include <algorithm>
   #include <cctype>
   #define num s-'0'
  
   using namespace std;
  
   struct edge
   {
       int u;
       int v;
       int cost;
   };
  
   const int MAX_V=;
   const int MAX_E=;
   const int INF=0x3f3f3f3f;
   int K,N, M, E, V;
   edge es[MAX_E];
   long long res;
   int par[MAX_V];
   int r[MAX_V];
  
   void kruskal();
   void init();
   int find(int);
   void unite(int, int);
   bool same(int, int);
  
   void read(int &x){
       char s;
       x=;
       bool flag=;
       while(!isdigit(s=getchar()))
           (s=='-')&&(flag=true);
       for(x=num;isdigit(s=getchar());x=x*+num);
       (flag)&&(x=-x);
   }
  
   void write(int x)
   {
       if(x<)
       {
           putchar('-');
           x=-x;
       }
       if(x>)
           write(x/);
       putchar(x%+'');
   }
  
   bool compare(const edge &e1, const edge &e2)
   {
       return e1.cost<e2.cost;
   }
  
   int main()
   {
       read(K);
       for (int j=; j<K; j++)
       {
           res=;
           read(N);read(M);read(E);V=N+M;
           for (int i=; i<E; i++)
           {
               read(es[i].u);read(es[i].v);read(es[i].cost);
               es[i].v+=N;es[i].cost=-es[i].cost;
           }
           kruskal();
           write(*(N+M)+res);
           putchar('\n');
       }
   }
  
   void init()
   {
       for (int i=; i<V; i++)
       {
           par[i]=i;
           r[i]=;
       }
   }
  
   int find(int x)
   {
       if (par[x]==x) return x;
       return par[x]=find(par[x]);
   }
  
   void unite(int x, int y)
   {
       x=find(x);
       y=find(y);
       if (x==y) return;
       if (r[x]<r[y]) par[x]=y;
       else
       {
           par[y]=x;
           if (r[x]==r[y]) r[x]++;
       }
   }
  
   bool same(int x, int y)
   {
       return (find(x)==find(y));
   }
  
   void kruskal()
   {
       sort(es, es+E, compare);
       init();
       for (int i=; i<E; i++)
       {
           edge e=es[i];
           if (!same(e.u, e.v))
           {
               unite(e.u, e.v);
               res+=e.cost;
           }
       }
   }