Connect the Cities
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12903 Accepted Submission(s):
3549
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.
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.
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.
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
prime算法
#include<stdio.h>
#include<string.h>
#define MAX 550
#define INF 0x3f3f3f
int city;
int visit[MAX],map[MAX][MAX],low[MAX];
void prime()
{
int j,i,min,mincost=0,next;
memset(visit,0,sizeof(visit));
for(i=1;i<=city;i++)
{
low[i]=map[1][i];
}
visit[1]=1;
for(i=1;i<city;i++)
{
min=INF;
for(j=1;j<=city;j++)
{
if(!visit[j]&&min>low[j])
{
next=j;
min=low[j];
}
}
if(min==INF)
{
printf("-1\n");
return ;
}
mincost+=min;
visit[next]=1;
for(j=1;j<=city;j++)
{
if(!visit[j]&&low[j]>map[next][j])
low[j]=map[next][j];
}
}
printf("%d\n",mincost);
}
int main()
{
int n,m,j,i,k,q,p,c,t,x,y,l;
int a[MAX];
scanf("%d",&n);
while(n--)
{
scanf("%d%d%d",&city,&m,&k);
for(i=1;i<=city;i++)
{
for(j=1;j<=city;j++)
{
if(i==j)
map[i][j]=0;
else
map[i][j]=INF;
}
}
for(i=1;i<=m;i++)
{
scanf("%d%d%d",&p,&q,&c);
if(map[p][q]>c)
map[p][q]=map[q][p]=c;
}
while(k--)
{
scanf("%d",&t);
for(l=1;l<=t;l++)
scanf("%d",&a[l]);
for(i=1;i<t;i++)
{
for(j=i+1;j<=t;j++)
{
map[a[i]][a[j]]=map[a[j]][a[i]]=0;
}
}
}
prime();
}
return 0;
}