HDU4496 D-City

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http://acm.hdu.edu.cn/showproblem.php?pid=4496

D-City

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1881 Accepted Submission(s): 690

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 5Hint
The graph given in sample input is a complete graph, that each pair of vertex has an edge connecting them, so there's only 1 connected block at first. The first 3 lines of output are 1s  because  after  deleting  the  first  3  edges  of  the  graph,  all  vertexes  still  connected together. But after deleting the first 4 edges of the graph, vertex 1 will be disconnected with other vertex, and it became an independent connected block. Continue deleting edges the disconnected blocks increased and finally it will became the number of vertex, so the last output should always be N.

Source

2013 ACM-ICPC吉林通化全国邀请赛——题目重现

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题目大意很容易读懂。解题思路：逆向并查集求联通分量。

如样例。当去除第m条边时，有5个联通分量。当去除第m-1条边时，有4个或更少的联通分量。以此类推

我们可以这么想。刚开始有5个散点，然后依次放第m条边 m-1条边 m-2条边看能形成多少个集合。大概意思如图

球集合的过程就是并查集代码如下;

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<vector>
#include<cmath>
#include<stdlib.h>
#include<iomanip>
#include<list>
#include<deque>
#include<map>
#include <stdio.h>
#include <queue>

#define maxn 10000+5
#define ull unsigned long long
#define ll long long
#define reP(i,n) for(i=1;i<=n;i++)
#define rep(i,n) for(i=0;i<n;i++)
#define cle(a) memset(a,0,sizeof(a))
#define mod 90001
#define PI 3.141592657

const ull inf = 1LL << 61;
const double eps=1e-5;

using namespace std;

bool cmp(int a,int b){
    return a>b;
}
struct node
{
    int s,e;
}edge[100005];
int fa[10005];
void init(int n)
{
   for(int i=0;i<n;i++)
    fa[i]=i;
}
int findfa(int x)
{
    if(x==fa[x])return x;
    else return fa[x]=findfa(fa[x]);
}
int Union(int x,int y)
{
    int fa_x=findfa(x);
    int fa_y=findfa(y);
    if(fa_x!=fa_y){fa[fa_x]=fa_y;return 1;}
    return 0;
}
int n,m;
int ans[100005];
int main()
{
     //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    while(~scanf("%d%d",&n,&m))
    {
        init(n);
        ans[m]=n;///去掉m跳边有n个分量
        for(int i=1;i<=m;i++)
            scanf("%d%d",&edge[i].s,&edge[i].e);
        for(int i=m;i>1;i--)///逆向并查集求来连通分量
        {
            int a=edge[i].s;
            int b=edge[i].e;
               if( Union(a,b))
                ans[i-1]=ans[i]-1;
                else ans[i-1]=ans[i];
        }
        for(int i=1;i<=m;i++)
            ///cout<<ans[i]<<endl;用cout输出会超时
        printf("%d\n",ans[i]);
    }
    return 0;
}

数量很大。用scanf()和printf()否则超时!`

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