HDU-4496 D-City （并查集）

D-City

http://acm.hdu.edu.cn/showproblem.php?pid=4496

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)

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. 

题目大意：给定一个无向图，删除前k条边(1<=k<=m)，求联通分量的个数？

感觉做过类似的。。。很快就想到倒着处理，就可以利用并查集写了

#include <cstdio>#include <cstring>using namespace std;const int MAXN=10005;int par[MAXN],ps,pe,n,m;int s[100005],e[100005],ans[100005];int getPar(int a) {    if(par[a]==a)        return a;    return par[a]=getPar(par[a]);}int main(){    while(2==scanf("%d%d",&n,&m)) {        for(int i=0;i<n;++i) {            par[i]=i;        }        for(int i=0;i<m;++i) {            scanf("%d%d",s+i,e+i);        }        ans[m]=n;        for(int i=m-1;i>0;--i) {            ps=getPar(s[i]);            pe=getPar(e[i]);            ans[i]=ans[i+1];            if(ps!=pe) {//如果他们不在同一个联通分量                --ans[i];                par[pe]=ps;            }        }        for(int i=1;i<=m;++i) {            printf("%d\n",ans[i]);        }    }return 0;}