D-City
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1315 Accepted Submission(s): 496
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.
Source 2013 ACM-ICPC吉林通化全国邀请赛――题目重现 解题报告 给定一个图，N个点，M条边，求按顺序删边后剩下的多少联通块。 一般我们都是用并查集来加边，删边也可以用并查集，删除第1条边等于加上m-1条边。。。
#include
#include
#include
#define M 100000+10 #define N 10000+10 using namespace std; int n,m; struct node { int u,v; } edge[M]; int B[N],ans[M]; int fine(int x) { if(B[x]!=x) B[x]=fine(B[x]); return B[x]; } int main() { int i; while(~scanf("%d%d",&n,&m)) { memset(ans,0,sizeof(ans)); memset(edge,0,sizeof(edge)); for(i=0;i
=0;i--) { ans[i]=sum; int xx=fine(edge[i].u); int yy=fine(edge[i].v); if(xx!=yy) { sum--; B[xx]=yy; } } for(i=0;i