lan adding key += spa appear eight family conn

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
Recommend lcy | We have carefully selected several similar problems for you: 1102 1301 1162 1198 1598
Statistic | Submit | Discuss | Note

難理解的就是最後那k行。開始的數字t表示有幾個城市。然後輸入t個城市，表示第一個城市和第二個連接，第二個和第三個連接。

。。

用kruskal算法超時的多提交兩次。

。當然也能夠用pri算法。。

不想寫。。

#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\n");
		else
		printf("%d\n",sum);
	}
	return 0;
}

hdu3371 Connect the Cities (MST)