1000搬桌子

Moving Tables

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9950 Accepted Submission(s): 3389

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
HDU1050

大概题意：输入一组数，搬桌子，走廊两边各有200个房间，要从s房间搬到t房间，走廊一次只能走一张桌子，找一种方案是用时最短

解题思路形成过程：通过参考课件，以及在网上搜索答案获得思路，解题的关键是贪心算法，主要思路是：

将每个房间之间的 走廊作为一个统计单位，当所有的办公桌都搬运完成之后，看看这段走廊到底需要占用多少次？然后统计所有的走廊被占用的最大值 max，这个值就是要单独安排的搬运次数，乘以10就是总的搬运时间。

我的疑问是：如果只是找使用最多次数的走廊，那题目给出的是否走廊重叠是干什么的？

我的收获是：学会了memset的用法，交换两个数的大小可以用swap，贪心算法很好用，但我还不会用，所以要多练习

解题过程：因为在定义数组时给a赋初值时在第一个for之前了，使数值没有及时清零，错了好几次

我的感想是：学好英语真的很重要，好多地方就是因为英语没学好理解错了，一直算不出来，因为并不完全是自己做出来的，尤其是思路根本没想到所以呵呵···以后吧，要是自己能写出来的话应该会很开心吧

AC代码：

 #include<iostream>
#include<algorithm>
#include<string>
#include<sstream>
#include<vector>
#include<numeric>
#include<cmath>
#include<cstdlib>
#include<set>
#include<map>
#include<queue>
#include<stdio.h>
#include<string.h>//尽管我后来知道了并不需要这么多头文件
using namespace std;
int main()
{
int a[201];
int T,N,s,t,m,i,j,b,c;
cin>>T;
for (c=0;c<T;c++)
{
cin>>N;
memset (a,0,sizeof (a));
m=0;
for(b=0;b<N;b++)
{
cin>>s>>t;
s=(s-1)/2;
t=(t-1)/2;
if (s>t)
swap (s,t);
for (i=s;i<=t;i++)
{
a[i]+=1;
}
}
for (j=0;j<200;j++)
{
if (a[j]>m)
m=a[j];
}
cout<<m*10<<endl;
}
return 0;
}