D-City
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 5582 Accepted Submission(s): 1957
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.
這個題正向求解很難想，我們不妨方向求解，将破壞線路，改為連接配接線路
通過并查集找出他們是不是屬于一個集合，如果不是一個集合說明破壞時增加了一個集合，應該再減去，（注意在哪個點上加）
//拆除線路判斷是否斷開不好判斷，
//是以反向連接配接判斷是否形成區域
#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#define MAXN 10005
using namespace std;
int operation [][];//記錄指令
int father[MAXN];
int sum[];//記錄結果
int num;
int n,m;
//并查集基本操作
int mfind(int x){
int r = x;
while(r!=father[r]){
r = father[r];
}
int i=x,j;
while(i!=r){
j = father[i];
father[i] = r;
i = j;
}
return r;
}
void connect(int x,int y){
int nx = mfind(x);
int ny = mfind(y);
if(nx!=ny){
father[ny] = nx;
num--;
}
}
int main()
{
while(~scanf("%d%d",&n,&m)){
num = n;
for(int i=;i<=n;i++){
father[i] = i;
}
for(int i=;i<m;i++){
scanf("%d%d",&operation[i][],&operation[i][]);
}
for(int i=m-;i>=;i--){
sum[i] = num;
connect(operation[i][],operation[i][]);
}
for(int i=;i<m;i++){
printf("%d\n",sum[i]);
}
}
return ;
}