POJ 3723 Conscription(最大生成树)
2016-09-20 18:18
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思路:比较显然的求个最大生成树然后10000*(n+m)-权值即可
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn = 50000+7;
struct Edge
{
int u,v,w;
}e[maxn];
bool cmp(Edge a,Edge b){return a.w>b.w;}
int fa[maxn];
int Find(int x){return fa[x]==x?x:fa[x]=Find(fa[x]);}
int n,m,k;
void init()
{
for(int i = 0;i<=n+m;i++)
fa[i]=i;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&n,&m,&k);
init();
for(int i = 1;i<=k;i++)
{
scanf("%d%d%d",&e[i].u,&e[i].v,&e[i].w);
e[i].u++;
e[i].v += (n+1);
}
int cnt = 0;
int sum = 0;
sort(e+1,e+1+k,cmp);
for(int i = 1;i<=k;i++)
{
int u = Find(e[i].u);
int v = Find(e[i].v);
if(u!=v)
{
fa[u]=v;
sum+=e[i].w;
if(++cnt>=n+m-1)
break;
}
}
printf("%d\n",10000*(n+m)-sum);
}
}
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
For each test case output the answer in a single line.
Sample Input
Sample Output
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn = 50000+7;
struct Edge
{
int u,v,w;
}e[maxn];
bool cmp(Edge a,Edge b){return a.w>b.w;}
int fa[maxn];
int Find(int x){return fa[x]==x?x:fa[x]=Find(fa[x]);}
int n,m,k;
void init()
{
for(int i = 0;i<=n+m;i++)
fa[i]=i;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d",&n,&m,&k);
init();
for(int i = 1;i<=k;i++)
{
scanf("%d%d%d",&e[i].u,&e[i].v,&e[i].w);
e[i].u++;
e[i].v += (n+1);
}
int cnt = 0;
int sum = 0;
sort(e+1,e+1+k,cmp);
for(int i = 1;i<=k;i++)
{
int u = Find(e[i].u);
int v = Find(e[i].v);
if(u!=v)
{
fa[u]=v;
sum+=e[i].w;
if(++cnt>=n+m-1)
break;
}
}
printf("%d\n",10000*(n+m)-sum);
}
}
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
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
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