题目：
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. 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
思路：
一开始的思路是第一组依次循环，如果遇到可以一起的将visit[j] = 1；
然而忽略了如：10 20；30 40； 35 45；30 40，和35,45都可以满足10 20，
可是35 45无法满足30 40，所以实际答案应该是2自己原先的代码得结果是1.
为了解决这个问题，可以将每组经过的过道都加1，这样过道经过数最多的就是组数
要最多的。
代码;
1 #include<stdio.h>
2 #include<iostream>
3 #include<algorithm>
4 #include<string.h>
5 using namespace std;
6
7 typedef struct{
8 int start;
9 int end;
10 }Table;
11 Table table[205];
12
13 int getmax(int x, int y){
14 return x > y ? x : y;
15 }
16 int getmin(int x, int y){
17 return x > y ? y : x;
18 }
19
20 int main(){
21 int t, n, i, j, visit[205], flag, num, max;
22 scanf("%d", &t);
23 while(t --){
24 scanf("%d", &n);
25 memset(visit, 0, sizeof(visit));
26 for(i = 0; i < n; i ++){
27 scanf("%d %d", &table[i].start, &table[i].end);
28 table[i].start = (table[i].start + 1) / 2;
29 table[i].end = (table[i].end + 1) / 2;
30 for(j = getmin(table[i].start, table[i].end); j <= getmax(table[i].start, table[i].end); j ++){
31 visit[j] ++;
32 }
33 }
34 sort(visit, visit + 205);
35 printf("%d\n", visit[204] * 10);
36 }
37 return 0;
38 }
奋进