HDU 4496 D-City(并查集）

D-City

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 2401    Accepted Submission(s): 841

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 manyconnected 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；题目比较短，题意应该看得出来。
思路；并查集。从后面开始加入，如果加入的两个点的父亲一样，对connected blocks的数量没影响， 如果不一样，说明两个connected blocks合在一起，数量减少1。
#include <stdio.h>#include <stdlib.h>#include <iostream>#include <algorithm>#include <cstring>#include <math.h>#include <vector>using namespace std;int a[100009], b[100009], s[100009], par[10009], ran[10009], ans[100009];void init(int n){    for(int i = 0; i < n; i++)    {        par[i]=i;        ran[i]=0;    }}int fin(int x){    if(par[x]==x) return x;    else return par[x]=fin(par[x]);}void unite(int x, int y){    x = fin(x);    y = fin(y);    if(x==y) return;    else    {        par[x]=y;        if(ran[x]==ran[y]) ran[x]++;    }}int main(){    int n, m;    while(~scanf("%d%d", &n, &m))    {        init(n);        for(int i=0; i<m; i++)        {            scanf("%d%d", &a[i], &b[i]);        }        memset(ans, 0, sizeof(ans));        ans[m-1]=n;        for(int i=m-1; i>0; i--)        {            if(fin(a[i])!=fin(b[i])) ans[i-1]=ans[i]-1;            else ans[i-1]=ans[i];            unite(a[i], b[i]);        }        for(int i=0; i<m; i++)            printf("%d\n", ans[i]);    }    return 0;}