BNUOJ 33895 D-City
D-City
This problem will be judged on HDU. Original ID: 4496
64-bit integer IO format: %I64d Java class name: Main
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
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
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
Source
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = ;
int uf[maxn],n,m;
int ans[maxn*];
int a[maxn*],b[maxn*];
int Find(int x){
if(x != uf[x])
uf[x] = Find(uf[x]);
return uf[x];
}
int main(){
int i,j,k;
while(~scanf("%d %d",&n,&m)){
for(i = ; i <= n; i++)
uf[i] = i;
for(i = ; i <= m; i++){
scanf("%d %d",a+i,b+i);
}
ans[m] = n;
for(i = m; i; i--){
int tx= Find(a[i]);
int ty = Find(b[i]);
uf[tx] = ty;
if(tx != ty){
ans[i-] = ans[i]-;
}else ans[i-] = ans[i];
}
for(i = ; i <= m; i++){
printf("%d\n",ans[i]);
}
}
return ;
}
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