2013
11-26

# 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.

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.

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

3
4
10 20
30 40
50 60
70 80
2
1 3
2 200
3
10 100
20 80
30 50 

10
20
30

2011-12-16 23:03:04

mark：其实就是最大重叠数。wa了一次，注意同一个点出来的两张桌子，如果一个往左，一个往右，不可以同时进行。

# include <stdio.h>
# include <stdlib.h>

typedef struct node{
int a, b ;
}node ;

node pt[500] ;

int min(int a, int b){return a<b?a:b;}
int max(int a, int b){return a>b?a:b;}
int cmp(const void *a, const void *b)
{
node *p = (node*)a, *q = (node*)b ;
if (p->a != q->a) return p->a - q->a ;
return q->b - p->b ;
}

int main ()
{
int i, T ;
int n, aa, bb, a, b ;
int ans, cur ;
scanf ("%d", &T) ;
while (T--)
{
scanf ("%d", &n);
for (i = 0; i < n ; i++)
{
scanf ("%d%d", &aa, &bb) ;
a = (min(aa,bb)+1) / 2 ;
b = (max(aa,bb)+1) / 2 ;
pt[i*2].a = a, pt[i*2].b = 1 ;
pt[i*2+1].a = b, pt[i*2+1].b = -1 ;
}
qsort(pt, n*2, sizeof(node), cmp) ;
ans = cur = 0 ;
for (i = 0 ; i < 2*n ; i++)
{
cur += pt[i].b ;
ans = max(ans, cur) ;
}
printf ("%d\n", ans*10) ;
}
return 0 ;
}

1. 约瑟夫也用说这么长……很成熟的一个问题了，分治的方法解起来o(n)就可以了，有兴趣可以看看具体数学的第一章，关于约瑟夫问题推导出了一系列的结论，很漂亮

2. 题目需要求解的是最小值，而且没有考虑可能存在环，比如
0 0 0 0 0
1 1 1 1 0
1 0 0 0 0
1 0 1 0 1
1 0 0 0 0
会陷入死循环

3. #include <stdio.h>
int main(void)
{
int arr[] = {10,20,30,40,50,60};
int *p=arr;
printf("%d,%d,",*p++,*++p);
printf("%d,%d,%d",*p,*p++,*++p);
return 0;
}

为什么是 20,20,50,40,50. 我觉得的应该是 20,20,40,40,50 . 谁能解释下？