Connect the Cities
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 13299 Accepted Submission(s): 3618
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.
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
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#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct graph{
int a;
int b;
int cost;
}G[ 25005];
int root[505];
int n,m;
int find(int i){
if(root[i]==i) return i;
return root[i]=find(root[i]);
}
void unio(int i,int j){
int t=find(i);
int k=find(j);
if(t<=k) root[k]=t;
else root[t]=k;
return ;
}
int cmp(graph u,graph v){
return u.cost<v.cost;
}
int shortest;
void kruskal(){
sort(G,G+m,cmp);
int i,t,k;
shortest=0;
for(i=0;i<m;++i){
t=find(G[i].a); k=find(G[i].b);
if(t!=k){
unio(t,k);
shortest+=G[i].cost;
}
}
return ;
}
int main(){
int T;
scanf("%d",&T);
while(T--){
int k,i,j,p,q,c,t,x,y;
scanf("%d%d%d",&n,&m,&k);
for(i=1;i<=n;++i) root[i] = i;
for(i=0;i<m;++i){
scanf("%d%d%d",&p,&q,&c);
G[i].a=p,G[i].b=q,G[i].cost=c;
}
for(i=1;i<=k;++i){
scanf("%d",&t);
scanf("%d%d",&x,&y);
unio(x,y);
for(j=3;j<=t;++j){
scanf("%d",&y);
unio(x,y);
}
}
kruskal();
int nu=0;
for(i=1;i<=n;++i){
if(find(i)==i) nu++;
}
if(nu>1) printf("-1
");
else printf("%d
",shortest);
}
return 0;
}