HDU 4496 并查集

Posted 2020-10-09 zzq

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

 

D-City

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 4802    Accepted Submission(s): 1690

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.

 

 

     大意是给出N点M边每次除去这条边之后询问此时有多少个联通块。

     一开始想的是统计每个点的入度一旦有点入度为零就加一，后来发现不对，对于一片树来说统计联通个数这样搞很乱而且不对。

     正解用并查集，逆向考虑，到i的时候拥有的边是i下面的，没有的是i上面的，所以我们从下至上将边添加进去，显然初始时联通快个数就是N，因为没边，当合并两棵树时-1即可，最后顺序输出。

  

 1 #include<cstring>
 2 #include<cstdio>
 3 #include<map>
 4 #include<iostream>
 5 using namespace std;
 6 struct Edge{int u,v,res;}e[100005];
 7 int f[10005];
 8 int getf(int v){return f[v]==v?v:f[v]=getf(f[v]);}
 9 int main()
10 {
11     int N,M,i,j,k;
12     while(cin>>N>>M){
13         for(i=0;i<N;++i) f[i]=i;
14         int u,v,ans=N;
15         for(i=1;i<=M;++i)
16         {
17             scanf("%d%d",&e[i].u,&e[i].v);
18         }
19         for(i=M;i>=1;--i)
20         {
21             e[i].res=ans;
22             int fu=getf(e[i].u),fv=getf(e[i].v);
23             if(fu!=fv){
24                 f[fu]=fv;
25                 ans--;
26             }
27 
28         }
29         for(i=1;i<=M;++i) printf("%d\n",e[i].res);
30     }
31     return 0;
32 }