图论之最小生成树-Conscription

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-dRMB. 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

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

这题是比较典型的最小生成树直接查找最多能省多少钱

#include<map>#include<set>#include<cmath>#include<queue>#include<bitset>#include<math.h>#include<vector>#include<string>#include<stdio.h>#include<cstring>#include<iostream>#include<algorithm>using namespace std;int f[20010];int rank1[20010];struct example{int x;int y;int d;}a[50010];int cmp(example a,example b){return a.d>b.d;}void init(){int i;for(i=0;i<=20010;i++){f[i]=i;rank1[i]=0;}}int find(int x){int i,j=x;while(j!=f[j])j=f[j];while(x!=j){i=f[x];f[x]=j;x=i;}return j;}void Union(int x,int y){int t1=find(x),t2=find(y);if(t1==t2)return ;if(rank1[t1]>rank1[t2])f[t2]=t1;else{f[t1]=t2;if(rank1[t1]==rank1[t2])rank1[t2]++;}}int main(){int all;int n,m,r;int i,j,k;scanf("%d",&all);while(all--){int money=0;scanf("%d%d%d",&n,&m,&r);for(i=0;i<r;i++){scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].d);a[i].y+=10000;}sort(a,a+r,cmp);init();for(i=0;i<r;i++){int a1=find(a[i].x);int a2=find(a[i].y);if(a1!=a2){money+=a[i].d;Union(a1,a2);}}money=(n+m)*10000-money;printf("%d\n",money);}return 0;}