http://acm.hdu.edu.cn/showproblem.php?pid=1050
题意：不占用通道就可以同时过，占用通道就不能同时过，问最短的时间全部搬完桌子。
贪心策略--每个房子前的通道用数组a[200]表示，占用通道就加+1；占用通道最大的数，就是最短的时间再乘以10；
需要注意的是房间与走廊的位置关系，就如图中所给的图片所示
相对门的两个房间占用同一处走廊，所以例如有1——3和4——6移动方式时，它们是共享了同一段走廊的，即3号房间门前的走廊。
分析过程有如下图所示
Moving Tables
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19535 Accepted Submission(s):
6678
Problem 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.
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
#include<iostream> #include<cstring> #include<cstdio> using namespace std; int a[205]; int main() { int T,n,m,t,q,k,i,x,max; scanf("%d",&T); while(T--) { memset(a,0,sizeof(a)); scanf("%d",&x); while(x--) { scanf("%d%d",&n,&m); if(n>m) { t=n; n=m; m=t; } if(n%2==0) q=n/2; else q=(n+1)/2; if(m%2==0) k=m/2; else k=(m+1)/2; for(i=q;i<=k;i++) a[i]=a[i]+1;; } max=0; for(i=1;i<=200;i++) { if(max<a[i]) max=a[i]; } printf("%d ",max*10); } return 0; }