HDU 4496 D-City（并查集）

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本文解析了一道关于并查集的基础题目，介绍了如何通过并查集算法解决城市连接问题，详细阐述了输入输出格式及样例解释，并提供了完整的代码实现。

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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.

并查集基础题，最终状态时必有N个连通分图，可以考虑从第N个开始倒着将边加入，记录每次加入边前的连通分图数，再顺序输出。

代码如下：



#include<stdio.h>
int city[100005][2],f[100005],ans[100005];
int getfather(int x){
    if(x!=f[x]) f[x]=getfather(f[x]);
    return f[x];
}
int main(){
    int n,m,i,a,b;
    int sum;
    while(scanf("%d %d",&n,&m)!=EOF){
        for(i=0;i<=n;++i) f[i]=i;
        for(i=0;i<m;++i){
            scanf("%d %d",&city[i][0],&city[i][1]);
            ans[i]=0;
        }
        sum=n;
        for(i=m-1;i>=0;--i){
            a=getfather(city[i][0]),b=getfather(city[i][1]);
            if(a==b){
                ans[i]=sum;
            }
            else{
                f[a]=b;///////
                ans[i]=sum--;
            }
        }
        for(i=0;i<m;++i) printf("%d\n",ans[i]);
    }
}