HDOJ 3371 Connect the Cities(最小生成树--prime 水)

Connect the Cities

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 13663    Accepted Submission(s): 3692


[align=left]Problem Description[/align]
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.  
 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
For each case, output the least money you need to take, if it’s impossible, just output -1.

 

[align=left]Sample Input[/align]

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

 

[align=left]Sample Output[/align]

1

ac代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#define MAXN 10100
#define INF 0x7fffffff
#define min(a,b) (a>b?b:a)
#define max(a,b) (a>b?a:b)
using namespace std;
int pri[MAXN][MAXN];
int v[MAXN];
int d[MAXN];
int dis[MAXN];
int n;
int sum;
void prime()
{
int i,j,k;
int min;
memset(v,0,sizeof(v));
for(i=1;i<=n;i++)
dis[i]=pri[1][i];
v[1]=1;
sum=0;
for(i=1;i<n;i++)
{
min=INF;
for(j=1;j<=n;j++)
{
if(v[j]==0&&dis[j]<min)
{
min=dis[j];
k=j;
}
}
if(min==INF)
break;
v[k]=1;
sum+=min;
for(j=1;j<=n;j++)
{
if(v[j]==0&&dis[j]>pri[k][j])
dis[j]=pri[k][j];
}
}
if(i!=n)
sum=-1;
}
int main()
{
int i,j,a,b,c,m,s;
int t,q,num,k;
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&n,&m,&k);
for(i=1;i<=n;i++)
{
pri[i][i]=0;
for(j=i+1;j<=n;j++)
pri[i][j]=pri[j][i]=INF;
}
for(i=0;i<m;i++)
{
scanf("%d%d%d",&a,&b,&c);
if(pri[a][b]>c)
pri[a][b]=pri[b][a]=c;
}
for(i=0;i<k;i++)
{
scanf("%d",&num);
for(j=0;j<num;j++)
{
scanf("%d",&d[j]);
}
for(j=0;j<num;j++)
for(q=j+1;q<num;q++)
pri[d[q]][d[j]]=pri[d[j]][d[q]]=0;
}
prime();
printf("%d\n",sum);
}
return 0;
}