Moving Tables
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
题目链接：
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1029
题意描述：
一共有400扇门，两侧各200扇，一个东西从一扇门搬运到另一扇门所需要的时间是10分钟，两侧门之间的过道不能并排移动两样东西，给你指定需要移动的物品，问最少要多久移动完。
解题思路：
先把这些物品存到结构体里，然后按a也就是物品出发的房间号排序，然后遍历后面要移动的物品，只要保证后面的物品b也就是物品终点房间号大于前面a物品的房间号就能同时移动，当然需要注意的是前一个物品终点的房间号与后一个物品的出发房间号是对门的时候需要b>a+1。
程序代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct data{
int a;
int b;
}s[510];
int book[510];
int cmp(data x,data y);
int main()
{
int T,n,i,j,sum,q,t;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%d%d",&s[i].a,&s[i].b);
if(s[i].a>s[i].b)
{
t=s[i].a;
s[i].a=s[i].b;
s[i].b=t;
}
}
sort(s,s+n,cmp);
memset(book,0,sizeof(book));
sum=0;
for(i=0;i<n;i++)
{
q=s[i].b;
if(book[i]==0)
{
sum++;
for(j=i+1;j<n;j++)
if(book[j]==0)
{
if(s[j].a%2==0&&s[j].a>q+1)
{
q=s[j].b;
book[j]=1;
}
if(s[j].a%2==1&&s[j].a>q)
{
q=s[j].b;
book[j]=1;
}
}
}
}
printf("%d\n",sum*10);
}
return 0;
}
int cmp(data x,data y)
{
if(x.a!=y.a)
return x.a<y.a;
return x.b<y.b;
}