hdu 3371(启发式合并的最小生成树)

Connect the Cities

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16070    Accepted Submission(s): 4177

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

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

 

题意：有n个城市,可以修其中的m条道路，然后还有k行,每一行有t个城市，这t个城市都已经连通。

题解：开始以为是普通的并查集，结果死活TLE，用了启发式合并才AC。而且就像很多人的题解里面说的用C++ AC不了。

启发式合并：启发式合并是为了解决合并过程中树退化成链的情况，用dep[i]表示根为i的树的最大深度，合并ra和rb时，采用最大深度小的向最大深度大的进行合 并，如果两棵树的最大深度一样，则随便选择一个作为根，并且将根的最大深度dep自增1，这样做的好处是在n次操作后，任何一棵集合树的最大深度都不会超过log(n)，所以使得查找的复杂度降为O( log(n) ）。

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;
const int N = 505;
const int M = 25000;
struct Edge
{
    int s,e,len;
} edge[M];
int father[N],n,m,k;
int dep[N];
int _find(int x)
{
    if(x==father[x])return x;
    return _find(father[x]);
}
int cmp(Edge a,Edge b)
{
    return a.len<b.len;
}
int kruskal(int m)
{
    sort(edge+1,edge+m+1,cmp);
    int cost = 0;
    for(int i=1; i<=m; i++)
    {
        int x  = _find(edge[i].s);
        int y  = _find(edge[i].e);
        if(x!=y)
        {
            if(dep[x]==dep[y])
            {
                father[x] = y;
                dep[y]++;
            }
            else if(dep[x]<dep[y])
            {
                father[x] = y;
            }
            else
            {
                father[y]=x;
            }
            cost += edge[i].len;
        }
    }
    return cost;
}
int main()
{
    int tcase;
    scanf("%d",&tcase);
    while(tcase--)
    {
        scanf("%d%d%d",&n,&m,&k);
        for(int i=1; i<=n; i++)
        {
            father[i] = i;
            dep[i] =0;
        }
        for(int i=1; i<=m; i++)
        {
            scanf("%d%d%d",&edge[i].s,&edge[i].e,&edge[i].len);
        }
        while(k--)
        {
            int t,a;
            scanf("%d%d",&t,&a);
            t--;
            while(t--)
            {
                int b;
                scanf("%d",&b);
                int x = _find(a);
                int y = _find(b);
                if(x!=y)
                {
                    if(dep[x]==dep[y])
                    {
                        father[x] = y;
                        dep[y]++;
                    }
                    else if(dep[x]<dep[y])
                    {
                        father[x] = y;
                    }
                    else
                    {
                        father[y]=x;
                    }
                }
            }
        }
        int ans = 0;
        int cost = kruskal(m);
        for(int i=1; i<=n; i++)
        {
            if(father[i]==i) ans++;
        }
        if(ans==1)printf("%d
",cost);
        else printf("-1
");
    }
}