Time Limit: 1000MS
Memory Limit: 65536K
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.
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
For each test case output the answer in a single line.
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
71071
54223
题意:组建一支军队,包括n个女孩m个男孩,每雇佣一个人需要花费10000元。若女孩x和男孩y有d关系,那么可以花费10000雇佣了x就可以花费10000-d来雇佣y。每个人只能使用一次关系。给出女孩和男孩的 r 组关系,求最少的花费。
解决:
方法1:最大生成树,ans=(n+m)*10000 - 最大生成树(森林)
方法2:最小生成树,每条边权值为10000-d
#include
#include
#include
#include
using namespace std;
#define maxn 50010
#define INF 0x3f3f3f3f
int t,n,m,r;//t样例n女m男r组关系
int x,y,d;//x与y有d关系
int ans,cnt;
int f[maxn];//父亲节点
struct node{
int x;
int y;
int d;
}people[maxn];
int cmp(node a,node b)
{
return a.d>b.d;
}
void init()
{
for(int i=1;ivoid add(int x,int y,int d)
{
people[cnt].x=x;
people[cnt].y=y;
people[cnt].d=d;
cnt++;
}
int find(int x)
{
if(x!=f[x]) f[x]=find(f[x]);
return f[x];
}
int kruskal()
{
sort(people,people+cnt,cmp);
for(int i=0;iint fx=find(people[i].x);
int fy=find(people[i].y);
if(fx!=fy)
{
f[fy]=fx;
ans-=people[i].d;
}
}
}
int main()
{
cin>>t;
while(t--)
{
cnt=0;
init();
cin>>n>>m>>r;//n女m男r组关系
while(r--)
{
cin>>x>>y>>d;//x与y有关系d
add(x,y+n,d);
}
ans=(n+m)*10000;
kruskal();
cout<return 0;
}