HDU 4496--D-City 【并查集删边】

D-City

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 2134    Accepted Submission(s): 751

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. 
 

 

题意：给出一个图，m条边，输出删除前i条边后该图的联通块的个数。

思路：从后往前操作，从后往前添加i条边等于添加完m条边后删掉前m-i条边，可知刚开始没有边，所以str[m]=n;

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int per[110000];int a[110000];int b[110000];int str[110000];int find(int x){    if(per[x] == -1) return x;    return per[x] = find(per[x]);}int main (){    int n,m;    while(scanf("%d %d", &n, &m)!=EOF){        memset(per, -1, sizeof(per));        for(int j = 1; j <= m; ++j)            scanf("%d %d", &a[j], &b[j]);        str[m] = n;        for(int i = m; i > 1; i--){            int fa = find(a[i]);            int fb = find(b[i]);            if(fa != fb){                per[fa] = fb;                str[i - 1] = str[i] - 1;            }            else str[i - 1] = str[i];        }        for(int i = 1; i <= m; ++i)            printf("%d\n",str[i]);    }    return 0;}