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.
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.
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.
逆向思维的重要性！
#include<cstdio>
#include<iostream>
#include<string.h>
#include<stack>
#define maxn 10005
using namespace std;
int pre[maxn];
int num;
void init()
{
for(int i=;i<maxn;i++) pre[i]=i;
}
int find(int x)
{
if(pre[x]==x) return x;
else return pre[x]=find(pre[x]);
}
void merge(int x,int y)
{
x=find(x);
y=find(y);
if(x!=y)
{
num--;
pre[y]=x;
}
}
int main()
{
cin.sync_with_stdio(false);
int n,m;
while(cin>>n>>m)
{
stack<int> fuck,fuck2;
int ret[m];
while(m--)
{
int x,y;
cin>>x>>y;
fuck.push(x),fuck.push(y);
}
init();
num=n;
int rett=;
while(!fuck.empty())
{
int yy=fuck.top();
fuck.pop();
int xx=fuck.top();
fuck.pop();
ret[rett++]=num;
merge(xx,yy);
}
for(int i=rett-;i>=;i--) cout<<ret[i]<<endl;
}
return;
}