hdu3371 Connect the Cities (MST)

Connect the Cities

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13722    Accepted Submission(s): 3711

Problem Description

In 2100, since the sea level rise, most of the cities disappear. Though some survived cities are still connected with others, but most of them become disconnected. The government wants to build some roads to connect all of these cities again, but they don’t want to take too much money.  

 

Input

The first line contains the number of test cases.
Each test case starts with three integers: n, m and k. n (3 <= n <=500) stands for the number of survived cities, m (0 <= m <= 25000) stands for the number of roads you can choose to connect the cities and k (0 <= k <= 100) stands for the number of still connected cities.
To make it easy, the cities are signed from 1 to n.
Then follow m lines, each contains three integers p, q and c (0 <= c <= 1000), means it takes c to connect p and q.
Then follow k lines, each line starts with an integer t (2 <= t <= n) stands for the number of this connected cities. Then t integers follow stands for the id of these cities.

 

Output

For each case, output the least money you need to take, if it’s impossible, just output -1.

 

Sample Input

1
6 4 3
1 4 2
2 6 1
2 3 5
3 4 33
2 1 2
2 1 3
3 4 5 6

 

Sample Output

1

 

Author

dandelion

 

Source

HDOJ Monthly Contest – 2010.04.04

 

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Statistic | Submit | Discuss | Note

难理解的就是最后那k行。開始的数字t表示有几个城市。然后输入t个城市，表示第一个城市和第二个连接，第二个和第三个连接。

。。

用kruskal算法超时的多提交两次。

。当然也能够用pri算法。。

不想写。。

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
struct node
{
	int a,b,cost;
}c[30000];
int fa[505];
void init(int n)
{
	for(int i=1;i<=n;i++)
	fa[i]=i;
}
bool cmp(node x,node y)
{
	return x.cost<y.cost;
}
int find(int x)
{
	if(fa[x]!=x) fa[x]=find(fa[x]);
	return fa[x];
}
int main()
{
	int n,k,m,ncase;
	scanf("%d",&ncase);
	while(ncase--)
	{
		scanf("%d %d %d",&n,&k,&m);
		init(n);
		for(int i=0;i<k;i++)
		scanf("%d %d %d",&c[i].a,&c[i].b,&c[i].cost);
		for(int i=1;i<=m;i++)
		{
			int x,pos,pos1;
			scanf("%d %d",&x,&pos);
			for(int j=1;j<x;j++)
			{
				scanf("%d",&pos1);
				c[k].a=pos,c[k].b=pos1,c[k].cost=0;
				pos=pos1;
				k++;
			}
		}
		sort(c,c+k,cmp);
		int sum=0;
		for(int i=0;i<k;i++)
		{
			int x=find(c[i].a);
			int y=find(c[i].b);
			if(x!=y)
			sum+=c[i].cost,fa[x]=y;
		}
		int count=0;
		for(int i=1;i<=n;i++)
		if(fa[i]==i)
		count++;
		if(count!=1)
		printf("-1
");
		else
		printf("%d
",sum);
	}
	return 0;
}