本来以为这题很难做。。。后来想了想发现不过是个贪心 ，因为每个过道的占用次数不论你怎么变换搬桌子顺序都不会变，所以 就把最大占用次数找出来 就知道最少耗时了
开一个cor[201] 数组 来存过道的占用次数 注意到房间和过道的对应关系是 （奇数+1）/2 偶数 /2 因而可以建立房间与过道的对应关系
题目及AC代码如下：
Moving Tables
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11544 Accepted Submission(s): 3944
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
1 #include<stdio.h> 2 #include<string.h> 3 #include<stdlib.h> 4 const int max_n=201; 5 6 int cmp(const void*a,const void *b) 7 { 8 int *a_=(int*)a; 9 int *b_=(int*)b; 10 return *b_-*a_; 11 } 12 13 14 int main(void) 15 { 16 int cor[max_n]; 17 int n; 18 scanf("%d",&n); 19 for(int i=0;i<n;i++) 20 { 21 int t; 22 memset(cor,0,sizeof(cor)); 23 scanf("%d",&t); 24 while(t--) 25 { 26 int a,b,temp; 27 int start_,end_; 28 scanf("%d%d",&a,&b); 29 if(a>b) {temp=a;a=b;b=temp;} //交换保证从小的移动到大的 30 start_=a%2==0?a/2:(a+1)/2; //犯低级错误~！！！！ 31 end_=b%2==0?b/2:(b+1)/2; 32 for(int oo=start_;oo<=end_;oo++) //每经过一次过道就利用率加一 33 cor[oo]++; 34 } 35 qsort(cor,max_n,sizeof(cor[0]),cmp); 36 printf("%d",cor[0]*10); 37 if(i<n-1) printf(" "); 38 } 39 }