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 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 3 + N -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 50Sample Output
10 20 30
题目大意
一层里面有400个房间，北边和南边各有200个房间，要从一个房间里面把一张桌子移动到另一个房间，需要占用这两个房间之间的所有走廊，每移动一个桌子需要10分钟，给出需要移动的桌子的数据，要求计算出最少需要多少分钟才能把所有桌子移动完。
解题思路
1和2 公用走廊1，3和4公用走廊2以此类推。并且从2到4要用走廊1和走廊2，挪1到2的同时不能挪2-4
.这里记录所有操作需要占用的走廊，占用次数最多的就是必须分成这么多组操作，所以就是答案。
代码
#include <cstring> #include <cstdio> #include <iostream> const int maxn=0x3f3f3f3f; using namespace std; int main() { int t; scanf("%d",&t); while(t--) { int n; int i,max=0; int a[210]={0}; scanf("%d",&n); while(n--) { int s,e; scanf("%d%d",&s,&e); if(s%2) s=s+1;//较小的奇数和较大的偶数占一个走廊 比如1和2 if(e%2) e=e+1; if(s>e) swap(s,e);//可能从大的房间搬到小的房间 for(i=s/2;i<=e/2;++i) { a[i]++; if(a[i]>max) max=a[i]; } } printf("%d\n",max*10); } return 0; }
