HDoj 4496 D-City（并查集）

D-City

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 2523    Accepted Submission(s): 883

Problem Description

Luxer is a really bad guy. He destroys everything he met.
One day Luxer went to D-city. D-city has N D-points and M D-lines. Each D-line connects exactly two D-points. Luxer will destroy all the D-lines. The mayor of D-city wants to know how many connected blocks of D-city left after Luxer destroying the first K D-lines in the input.
Two points are in the same connected blocks if and only if they connect to each other directly or indirectly.

 

Input

First line of the input contains two integers N and M.
Then following M lines each containing 2 space-separated integers u and v, which denotes an D-line.
Constraints:
0 < N <= 10000
0 < M <= 100000
0 <= u, v < N.

 

Output

Output M lines, the ith line is the answer after deleting the first i edges in the input.

 

Sample Input

   
   
   
   

    
    
    
    5 10 
0 1 
1 2 
1 3 
1 4 
0 2 
2 3 
0 4 
0 3 
3 4 
2 4
   
   
   
   

 

Sample Output

  
  
  
  

   
   
   
   1 
1 
1 
2 
2 
2 
2 
3 
4 
5


考并查集的建法。。
从前向后减=从后向前建


   
   
   
   
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int init[10010],ans[100010];int l1[100010],l2[100010];
int fi(int r)
{
    return r==init[r]?r:init[r]=fi(init[r]);
}
void mer(int a,int b)
{
//    int x=fi(a);
//    int y=fi(b);
    if(a>b)
        init[a]=b;
    else
        init[b]=a;
}
int main()
{
    int n,m,i,j,k,a,b;
    ios::sync_with_stdio(false);
    while(cin>>n>>m)
    {
        for(i=0;i<n;i++)
        init[i]=i;
        for(i=0;i<m;i++)
        {
            cin>>a>>b;
            l1[i]=a;//分别保存与数组中
            l2[i]=b;
        }
        ans[m]=n;//减到一条路没有必然是剩下n个城市 
        for(i=m-1;i>=0;i--)
        {
            if(fi(l1[i])!=fi(l2[i]))
            {
                mer(fi(l1[i]),fi(l2[i]));<span id="transmark"></span>
                ans[i]=ans[i+1]-1;
            }
            else
                ans[i]=ans[i+1];
        }
        for(i=1;i<=m;i++)
        {
            printf("%d\n",ans[i]);
        }
    }
    return 0;
}