ZOJ 1029 Moving Tables（贪心算法）
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
Output for the Sample Input
10
20
30
题解：
（题意是说：有400个房间南北两向各200个房间，装修需要搬运办公桌（桌子很大走廊只能通过一张办公桌）一张桌子从一个房间到里另外一个房间最多用10min。在这个时间内走廊一直被占用，每十分钟内只要不是同一个走廊均可以移动办公桌。每个房间，只有一张桌子进出，现在需要找出一种方案，提高搬运效率，尽快完成》》》》》》》》》》》）
刚开始把这道题想错了，错当成区间问题（会场安排）那个题，（能尽量多的安排会场），后来提交的结果是错的，换了个思路
房子是对着的，每两个房间之间是一小段的走廊。共四百个房间，分为
——————————————————————————————————————————————
1、2 | 3，4 | 5，6 | 7，8 |：：：：：：：：：：：：：：：：：：399，400
0 | 1 | 2 | 3 199
______________________________________________________________________________________
共分为200个走廊段（0——199）如果走廊段没有重叠的话则一次就可以搬运完（先求出最少的占用走廊的时间然后在乘10）。
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm>
using namespace std;
int main()
{
int mov[200];
int t;
scanf("%d",&t);
while(t--)
{
memset(mov,0,sizeof(mov));
int from,to;
int n;
scanf("%d",&n);
for(int i=0;i<n;i++)
{
scanf("%d %d",&from,&to);
from=(from-1)/2; //起点所在的走廊段
to=(to-1)/2; //终点所在的走廊段
int t;
if(from>to) //若起点所在的走廊段大于终点的走廊段（交换）
{
t=from;
from=to;
to=t;
}
for(int j=from;j<=to;j++)
{
mov[j]++; //在同一时间所在的走廊段（如果每个走廊段均为1则只需要十分钟若）
//(大于1则表明有走廊段同是被占用需等待后再搬运)。
}
}
int maxx=0;
for(int i=0;i<200;i++)
{
if(mov[i]>maxx)
maxx=mov[i]; //重合走廊段最大的那个数为最终需要的最少时间。
}
printf("%d\n",maxx*10);
}
return 0;
}