Connect the Cities
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9941 Accepted Submission(s): 2827
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
Mean:
好久没刷水题了。。。
给你一个有向图，其中某些边是连接的，现在要你找一棵最小生成树。
analyse:
先用并查集来合并已经连通的边，然后在使用kruskal来求最小生成树，注意这儿有一个剪枝，因为这题的边很多，所以我们不需要全部判断，根据最小生成树的性质可知，我们只需要找出n-1条边即可。
Time complexity：O(n)
Source code：
//Memory Time
// 5644K 241MS
//by : Snarl_jsb
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<iomanip>
#include<string>
#include<climits>
#include<cmath>
#define MAXV 520
#define MAXE 25010
#define LL long long
using namespace std;
int T,n,m,k;
struct Node
{
int s,e,val;
} graph[MAXE];
int father[MAXV];
void scan(int& x){
x = 0;
char c = getchar ();
while (!(c>='0' && c<='9' || c=='-')) c = getchar ();
while (c >= '0' && c <= '9'){
x = x * 10 + c - '0';
c = getchar ();
}
}
int Find(int x)
{
return x==father[x]?x:father[x]=Find(father[x]);
}
bool cmp(Node a,Node b)
{
return a.val<b.val;
}
int kruskal()
{
int ans=0;
int cnt=0;
sort(graph+1,graph+1+m,cmp);
for(int i=1;i<=n;i++)
if(father[i]==i)
cnt++;
for(int i=1;i<=m&&cnt>1;i++)
{
int x=Find(graph[i].s);
int y=Find(graph[i].e);
if(x!=y)
{
father[x]=y;
ans+=graph[i].val;
cnt--;
}
}
if(cnt==1)
return ans;
else return -1;
}
int main()
{
// freopen("cin.txt","r",stdin);
// freopen("cout.txt","w",stdout);
scan(T);
while(T--)
{
scan(n);
scan(m);
scan(k);
for(int i=1;i<=m;i++)
{
scan(graph[i].s);
scan(graph[i].e);
scan(graph[i].val);
}
for(int i=0;i<=n;i++)
father[i]=i;
int cnt,fa,son;
while(k--)
{
scan(cnt);
scan(fa);
fa=Find(fa);
while(--cnt)
{
scan(son);
father[Find(son)]=fa;
}
}
int mincost=kruskal();
printf("%d\n",mincost);
}
return 0;
}
