ZOJ 3328 Wu Xing

Wu Xing Time Limit: 1 Second      Memory Limit: 32768 KB

Introduction

The Wu Xing, or the Five Movements, Five Phases or Five Steps/Stages, are chiefly an ancient mnemonic device, in many traditional Chinese fields.

The doctrine of five phases describes two cycles, a generating or creation cycle, also known as "mother-son", and an overcoming or destruction cycle, also known as "grandfather-nephew", of interactions between the phases.

Generating:

• Wood feeds Fire;
• Fire creates Earth (ash);
• Earth bears Metal;
• Metal carries Water (as in a bucket or tap, or water condenses on metal);
• Water nourishes Wood.

Overcoming:

• Wood parts Earth (such as roots) (or Trees can prevent soil erosion );
• Earth absorbs (or muddies) Water (or Earth dam control the water);
• Water quenches Fire;
• Fire melts Metal;
• Metal chops Wood.

With the two types of interactions in the above graph, any two nodes are connected by an edge.

Problem

In a graph with N nodes, to ensure that any two nodes are connected by at least one edge, how many types of interactions are required at least? Here a type of interaction should have the following properties:

• It can be represented as a set of directed edges in the graph.
• For each type of interaction, there should be one and only one edge starting at each node.
• For each type of interaction, there should be one and only one edge ending at each node.
• The interactions are made up of cycles, i.e. starting from an arbitrary node and following the edges with the same type of interaction, you can always reach the starting node after several steps.

Input

For each test case, there's a line with an integer N (3 <= N < 1,000,000), the number of nodes in the graph.

N = 0 indicates the end of input.

Output

For each test case, output a line with the number of interactions that are required at least.

Sample Input

5
0

Sample Output

2

Reference

http://en.wikipedia.org/wiki/Wu_Xing

题目大意：

1.这是一个有向边集；

2.只有一个起始边；

3.只有一个终点边；

4.可以组成一个环。

注意：这儿有个“坑”（只是本人被坑了），这个环是将所有点连接起来的，因为每个环要贯穿所有点，而不仅仅是最后所有环两点连通，未懂题意得情况下走的队友思路，呜呼了……

思路：所有点直接连通，需要  n*(n-1)/2条边，每个环需要n个边（多个环可能共用一条边），所以环的个数至少为  (n-1)/2，如果n-1是偶数，直接得出答案；如果n-1是奇数，剩下的边肯定又可以构成一个环，是  (n-1)/2+1（此处不好理解）。那么程序中就可写为    n/2。

代码：

#include<iostream>
using namespace std;
int main()
{
    int n;
    while(~scanf("%d",&n)&&n)
    {
        printf("%d\n",n/2);
    }
}