浙江财经大学
信息管理与工程学院

HDU 3371 Connect the Cities(Kruskal)

本文由 Ocrosoft 于 2016-07-31 15:47:04 发表

Connect the Cities

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

Total Submission(s): 17118    Accepted Submission(s): 4324

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
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
 

Solution

题意:还是连通村子,第一行是数据组数,第二行n,m,k,表示村子数,道路数,已连通数据行数;接下来m行,再然后k行,每行第一个表示已连通的村子数t,后面跟着t个数,表示这些村子间互相连通。

给出的数据表示,有可能两个村子间有多条路,不过这对kruskal没有影响。后面已连通的处理可以读取第一个,把这一行后面的村子都作为这个村子的子节点。

注:该代码C++超时,需要用G++提交。

#include <set>
#include <map>
#include <list>
#include <cmath>
#include <stack>
#include <queue>
#include <ctime>
#include <string>
#include <cstdio>
#include <vector>
#include <cctype>
#include <climits>
#include <sstream>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define ms(a) memset(a,0,sizeof(a))
#define INF INT_MAX
#define MAXN 500+10
using namespace std;
int n,m,k;
int pre[MAXN];
struct Edge
{
    int u,v,w;
    bool operator <(const Edge e) const
    {
        return w<e.w;
    }
} edge[25010];
int find(int x)
{
    if(pre[x]!=x)pre[x]=find(pre[x]);
    return pre[x];
}
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        cin>>n>>m>>k;
        for(int i=1; i<=n; i++)pre[i]=i;
        for(int i=1; i<=m; i++)
            scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].w);
        for(int i=1; i<=k; i++)
        {
            int t;
            scanf("%d",&t);
            int a,b;
            scanf("%d",&a);
            t--;
            while(t--)
            {
                scanf("%d",&b);
                pre[find(b)]=pre[find(a)];
                //a=b;
            }
        }
        int sum=0,cnt=0;
        sort(edge+1,edge+m+1);
        for(int i=1;i<=m;i++)
        {
            int x=find(edge[i].u),y=find(edge[i].v);
            if(x!=y)sum+=edge[i].w,pre[x]=y;
        }
        for(int i=1;i<=n;i++)if(pre[i]==i)cnt++;
        if(cnt==1)printf("%d\n",sum);
        else printf("-1\n");
    }
    return 0;
}

 

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