A - Ice_cream's world I

本文主要是介绍A - Ice_cream's world I，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

题目：

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 10
0 1
1 2
1 3
2 4
3 4
0 5
5 6
6 7
3 6
4 7

Sample Output

3

题意：

ice_cream的世界是一个富裕的国家，它拥有许多肥沃的土地。 今天，冰淇淋女王想要将土地授予勤奋的ACMers。 所以有一些瞭望塔被建立起来，并且建造了瞭望塔之间的墙壁，以便划分冰淇淋的世界，没有墙壁交叉。 一个围墙包围的土地必须只给一个ACMer，求出最多可以分到土地的ACMer的数量。

思路：

用并查集，然后判断环，一个环就是一块地；

代码如下：

#include<stdio.h>
#include<string.h>int n,m,a,b;
int f[1001];void init()//初始化，自己的祖宗是自己；
{for(int i=0; i<n; i++)f[i]=i;return ;
}int getf(int v)//查询祖宗；
{if(f[v]==v)return v;elsereturn f[v]=getf(f[v]);
}int merge(int v,int u)//判断
{int t1=getf(v);int t2=getf(u);if(t1==t2)//同一个祖宗，有一个环，就有一块地；return 1;f[t2]=t1;return 0;
}int main()
{while(~scanf("%d%d",&n,&m)){init();int i,sum=0;for(i=0; i<m; i++){scanf("%d%d",&a,&b);sum+=merge(a,b);//统计；}printf("%d\n",sum);}return 0;
}

 

这篇关于A - Ice_cream's world I的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！

---