HDOJ 2120 Ice_cream's world I

Ice_cream's world I

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 974    Accepted Submission(s): 566

Problem Description

ice_cream's world is a rich country, it has many fertile lands. Today, the queen of ice_cream wants award land to diligent ACMers. So there are some watchtowers are set up, and wall between watchtowers be build, in order to partition the ice_cream’s world. But how many ACMers at most can be awarded by the queen is a big problem. One wall-surrounded land must be given to only one ACMer and no walls are crossed, if you can help the queen solve this problem, you will be get a land.

 

Input

In the case, first two integers N, M (N<=1000, M<=10000) is represent the number of watchtower and the number of wall. The watchtower numbered from 0 to N-1. Next following M lines, every line contain two integers A, B mean between A and B has a wall(A and B are distinct). Terminate by end of file.

 

Output

Output the maximum number of ACMers who will be awarded.
One answer one line.

 

Sample Input

8 100 11 21 32 43 40 55 66 73 64 7

 

Sample Output

3

 

Author

Wiskey

 

Source

HDU 2007-10 Programming Contest_WarmUp 

嗯……虽然把题看懂了，但是，成环的代码是硬记下来的。所以刚上手的时候有种没有头绪的感觉。

但是，并查集，意外的好理解啊。

2016.8.1

成环我会了！！并查集也明白更多了！！！另外这题看的时候又差点没看懂，补个翻译。

题意：有一个女王，她拥有一片土地。土地上有很多瞭望塔。现在在不同的瞭望塔之间砌上墙（把两座塔作为两点的话，墙就是他们两个的连线）。问：被墙体分割后，一共有多少块小土地？

#include<stdio.h>int per[1001];int count;int Find(int r){while(r!=per[r])r=per[r];return r;}void Merge(int x,int y){int fx=Find(x);int fy=Find(y);if(fx!=fy)per[fx]=fy;elsecount++;}int main(){int n,m,a,b,i;while(~scanf("%d%d",&n,&m)){count=0;for(i=0;i<n;i++)per[i]=i;for(i=1;i<=m;i++){scanf("%d%d",&a,&b);Merge(a,b);}printf("%d\n",count);}return 0;}