HDU-1050 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. 

 

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

题意：现在有400个房间，分布在走廊的两侧，现要将桌子从一个房间移动到另一个房间，走廊的宽度只能允许一个桌子通过。给出t，

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <stack>
#include<math.h>

using namespace std;

const int maxn = 210;
typedef long long ll;
int a[maxn];
int t, n;
int main()
{
    int st, ed,k;
    cin >> t;
    while (t--)
    {
        memset(a, 0, sizeof a);
        cin >> n;
        while (n--)
        {
            cin >> st >> ed;
            if (st > ed)k = st, st = ed, ed = k;
            if (st % 2 == 1)st += 1;
            if (ed % 2 == 1)ed += 1;
            for (int i = st/2; i <= ed/2; ++i)a[i] ++;//求出每组移动所经过的房间数，因为是两侧所以要除2
        }
        int maxx = 0;
        for (int i = 1; i < 202; ++i)maxx = max(maxx, a[i]);//求出最大数
        cout << maxx * 10 << endl;
    }
    return 0;
}