hdu 4496 D-City 并查集

D-City

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

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

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

HINT

题意

给你个图，他不断删边，然后问你联通块的个数

题解：

倒着做，然后直接并查集就好了

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define test freopen("test.txt","r",stdin)  
#define maxn 200001
#define mod 10007
#define eps 1e-9
int Num;
char CH[20];
const int inf=0x3f3f3f3f;
const ll infll = 0x3f3f3f3f3f3f3f3fLL;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline void P(int x)
{
    Num=0;if(!x){putchar('0');puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
//**************************************************************************************

int fa[maxn];
int fi(int x)    
{    
    return fa[x]==x?x:fa[x]=fi(fa[x]);    
}    
  
bool merge(int x,int y)    
{    
    int fx=fi(x),fy=fi(y);    
    if(fx!=fy)
    {  
        fa[fx]=fy;  
        return true;  
    }  
    return false;  
}  

int ans[maxn];
int a[maxn],b[maxn];
int main()
{
    //test;
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF){
    for(int i=0;i<=n;i++)
        fa[i]=i;
    for(int i=1;i<=m;i++)
        a[i]=read(),b[i]=read();
    int tot=n;
    for(int i=m;i>=1;i--)
    {
        ans[i]=tot;
        int q=fi(a[i]),p=fi(b[i]);
        if(merge(q,p))
            tot--;
    }
    for(int i=1;i<=m;i++)
        cout<<ans[i]<<endl;
    }
}