HDOJ 4496 D-City(并查集变形，逆序 删边)



D-City

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 2656    Accepted Submission(s): 928

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：市长要在魔王破坏道路后修复道路，以确保每个地区畅通。给出n个城市，m条道路。魔王破坏的道路按序给 出，规定直接连通或者间接连通的区域为同一片地区。问在魔王破坏到k条路时，城市被分成了几个区域。依次输出。

题解： 这是一个并查集删边问题，在连接好的图中按照题意删去边径，判断剩下的联通区个数。处理时用逆序建图

的方法，从魔王最后破坏的路开始合并地区。

具体代码如下：

#include<cstdio>#include<cstring>#define max 100010int set[10010],count;int ans[max],x[max],y[max];int find(int x){int r=x;int t;while(r!=set[r])r=set[r];while(r!=x){t=set[x];set[x]=r;x=t;}return r;}void merge(int a,int b,int k){int fa=find(a);int fb=find(b);if(fa!=fb){count--;//连通之前没有连通地区后 set[fa]=fb;}ans[k]=count;}int main(){int n,m,i;while(scanf("%d%d",&n,&m)!=EOF){for(i=0;i<n;++i)    set[i]=i;for(i=0;i<m;++i)scanf("%d%d",&x[i],&y[i]);ans[m]=n;count=n;for(i=m-1;i>0;i--)   merge(x[i],y[i],i);for(i=1;i<=m;++i)   printf("%d\n",ans[i]);}return 0;}