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.

Table moving
Reason
Possible
( room 30 to room 50) and (room 60 to room 90)
no part of corridor is shared
(room 11 to room 12) and (room 14 to room 13)
no part of corridor is shared
Impossible
(room 20 to room 40) and (room 31 to room 80)
corridor in front of room 31 to room 40 is shared
(room 1 to room 4) and (room 3 to room 6)
corridor in front of room 3 is shared
(room 2 to room 8) and (room 7 to room 10)
corridor in front of room 7 is shared

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.

输入
The input consists of T test cases. The number of test cases ( T) is given in the first line of the input file. 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.

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

样例输入
3
4
10 20
30 40
50 60
70 80
2
1 3
2 200
3
10 100
20 80
30 50
样例输出
10
20
30

题目不难，英文有些难以理解。
题目意思就是从两个不同的房间移桌子，移动桌子需要占用房间之间的过道，占用一次需要10分钟，问至少需要多少时间。
以最后一次输入为例
用数组a[200]初始化为0
3
10 100 经过 5-50过道 这过道的值都+1
20 80 经过 10-80过道

这过道的值都+1
30 50

经过 15-25过道 这过道的值都+1
扫一遍 寻找 a[200]中的最大值 发现 15-25 这10个数都是3 最大 即 15-25这个过道必须得被占用3次
用 3*10（time）就可得出正确结论

#include<stdio.h>
int main()
{
    int x,s,n,t,i,temp;
    scanf("%d",&x);
    while(x--)
    {
        int ans[300]={0};
        //memset(ans,0,sizeof(ans));另一种初始化数组值为0的方法
        scanf("%d",&n);
        while(n--)
        {
            scanf("%d %d",&s,&t);
            if(s>t)
            {
                temp=t;t=s;s=temp;
            }
            for(i=(s+1)/2;i<=(t+1)/2;i++)
            {
                ans[i]++;
            }
        }
        int max=0;
        for(i=1;i<=200;i++)
        {
            if(max<ans[i])
            {
                max=ans[i];
            }
        }
        printf("%d\n",max*10);
    }
} 

memset(a,0,sizeof(a))意思就是将a数组所有值初始为0；
这里初始值只能是0或-1；
因为memset是对每个字节赋值，而int有4字节(32位)
比如这样,memset(a,1,sizeof(a));
则a中的每个元素都被赋值成为2进制数为00000001000000010000000100000001的数
转换成10进制就是16843009
所以，一般用memset对数组赋0或-1，赋其他的值就要用循环来实现。
用int a[10]={0};也可以初始化所有值为0；
int a[10]={10}；意思为a[0]的值为10，其他值为0；