poj3723 kruskal

 

 

如题：http://poj.org/problem?id=3723

 

Conscription

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8418 Accepted: 2930

Description

Windy has a country, and he wants to build an army to protect his country. He has picked upN 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 andR.
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

25 5 84 3 68311 3 45830 0 65920 1 30633 3 49751 3 20494 2 21042 2 7815 5 102 4 98203 2 62363 1 88642 4 83262 0 51562 0 14634 1 24390 4 43733 4 88892 4 3133

Sample Output

7107154223

Source

POJ Monthly Contest – 2009.04.05, windy7926778

 

 

 

很多题都是最基础算法的延伸，要注意发现，找准思路。

 

这一题就是如果编号x的女和y的男兵认识，就可以优惠，优惠价值是10000-已经招募的人和自己亲密度的最大值。

 

思路：构建无向图，1-N是女兵，N+y是男兵，当他们有一腿，就用权值连起来，这样按题目要求，应该取出一颗最大生成树，再用(M+N)*10000-最大生成树权值和。

还可以对边权取负，求最小生成树，结果是（M+N）*10000+最小生成树权值和。

 

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
#define MAXE 100005
#define MAXV 20005
struct edge
{
 int u,v,cost;
};

edge G[MAXE];
int f[MAXV];
int T,N,M,R;

int cmp(edge &a,edge &b)
{
 return a.cost<b.cost;
}

void init()
{
 int i;
 for(i=0;i<MAXV;i++)
  f[i]=i;
}
int find(int x)
{
 if(x==f[x])
  return x;
 else
  return f[x]=find(f[x]);
}
void unite(int a,int b)
{
 int fa=find(a);
 int fb=find(b);
 if(fa!=fb)
  f[fa]=fb;
}
int same(int a,int b)
{
 return find(a)==find(b);
}
int kruskal()
{
 init();
 sort(G,G+R,cmp);
 int i;
 int res=0;
 for(i=0;i<R;i++)
 {
  if(!same(G[i].u,G[i].v))
  {
   res+=G[i].cost;
   unite(G[i].u,G[i].v);
  }
 }
 return res;
}
int main()
{
 //freopen("C:\\1.txt","r",stdin);
 scanf("%d",&T);
 while(T--)
 {
  memset(G,0,sizeof(G));
  scanf("%d%d%d",&N,&M,&R);
  int i;
  for(i=0;i<R;i++)
  {
   int a,b,c;
   scanf("%d%d%d",&a,&b,&c);
   G[i].u=a;
   G[i].v=M+b;
   G[i].cost=-c;
  }
  printf("%d\n",10000*(M+N)+kruskal());
 }
 return 0;
}