贪心算法，顾名思义是在程序设计过程中，在得到问题的解过程中，每一步采取的策略都为当前的局部最优，当这样进行下去确定最后得到的解为全局最优解，则贪心算法有效。

贪心算法一般的基本步骤如下：

1.确定每一步的局部最优解可以构成全局最优

2.从问题的初始状态开始得到问题的初始解

3.接下来每一步都选择在某种条件下当前最优的解，循环递进，缩减问题规模

4.综合得到的局部最优解，得到全局最优解

例如hdoj中的

Moving Tables

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11505    Accepted Submission(s): 3931

 

Problem Description

The famous ACM (Advanced Computer Maker) Company has rented a floor of a building whose shape is in the following figure.
 

The floor has 200 rooms each on the north side and south side along the corridor. Recently the Company made a plan to reform its system. The reform includes moving a lot of tables between rooms. Because the corridor is narrow and all the tables are big, only one table can pass through the corridor. Some plan is needed to make the moving efficient. The manager figured out the following plan: Moving a table from a room to another room can be done within 10 minutes. When moving a table from room i to room j, the part of the corridor between the front of room i and the front of room j is used. So, during each 10 minutes, several moving between two rooms not sharing the same part of the corridor will be done simultaneously. To make it clear the manager illustrated the possible cases and impossible cases of simultaneous moving.
 

For each room, at most one table will be either moved in or moved out. Now, the manager seeks out a method to minimize the time to move all the tables. Your job is to write a program to solve the manager’s problem.

Input

The input consists of T test cases. The number of test cases ) (T is given in the first line of the input. Each test case begins with a line containing an integer N , 1<=N<=200 , that represents the number of tables to move. Each of the following N lines contains two positive integers s and t, representing that a table is to move from room number s to room number t (each room number appears at most once in the N lines). From the N+3-rd line, the remaining test cases are listed in the same manner as above.

Output

The output should contain the minimum time in minutes to complete the moving, one per line.

Sample Input

3 4 10 20 30 40 50 60 70 80 2 1 3 2 200 3 10 100 20 80 30 50

Sample Output

10 20 30

Source

Asia 2001, Taejon (South Korea)

思路如下：

1.使用数组标记搬运桌子的路径的开始到结尾经过路径

2.从走廊的最开始位置出发，循环比较得出当前路径标记的最大值

3.循环结束后综合了每次循环得到的局部最优值得到搬运路径的最大的重叠，则为问题的解

代码如下：

#include
#include
using namespace std;int rooms[201]={0};int main(){int T,N,begin,end,max;scanf("%d",&T);while(T--){max=0;for(int i=0; i<201; i++)rooms[i]=0;scanf("%d",&N);for(int j=0; jend)swap(begin,end);for(int k=(begin+1)/2; k<=(end+1)/2; k++)rooms[k]++;}for(int g=0; g<201; g++)if(rooms[g]>max)max=rooms[g];printf("%d\n",max*10);}
}