HDOJ 4496 D-City (反向思维并查集)

D-City

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 2887    Accepted Submission(s): 1020

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

Hint

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. 题意：给你一个图，n个点，m条边，一条一条边的删除，求有多少个联通块 思路：正常并查集是一条一条边的加，然后判断联通块，但这个刚好反过来，那么直接储存反过来记录查询就行了 ac代码：  

#include
#include
#include
#include
#include
#include
#include
#include
#include
#define MAXN 1010000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
using namespace std;
int gcd(int a,int b){return b?gcd(b,a%b):a;}
LL powmod(LL a,LL b,LL MOD){LL ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
//head
struct s
{
	int a,b;
}p[MAXN];
int pri[MAXN];
int ans[MAXN];
int find(int x)
{
	int r=x;
	while(r!=pri[r])
	r=pri[r];
	int i,j=x;
	while(j!=r)
	{
		i=pri[j];
		pri[j]=r;
		j=i;
	}
	return r;
}
int main()
{
	int n,m,i;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(i=0;i=0;i--)
		{
			int nx=find(p[i].a);
			int ny=find(p[i].b);
			if(nx!=ny)
			{
				pri[ny]=nx;
				cnt--;
				ans[i]=cnt+1;
			}
			else
			ans[i]=cnt;
		}
		for(i=0;i