HDU 1050 Moving Tables

Moving Tables

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13571    Accepted Submission(s): 4656

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 planis 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

 

此题也是简单的贪心题目，用数学方法就是找众数。可以让每次重复的地方加一，最后找到最大的就可以。另外还需要注意一下：1和2   3和4  可以看成一个。因而用

x=(x-1)/2; y=(y-1)/2;

 

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int main()
{
    int cas,n,s[200],x,y,t;
    scanf("%d",&cas);
    while(cas--){
        memset(s,0,sizeof(s));
        scanf("%d",&n);
        for(int i=0;i<n;i++){
            scanf("%d%d",&x,&y);
            x=(x-1)/2;y=(y-1)/2;
            if(x>y){
                t=x;x=y;y=t;
            }
            for(int j=x;j<=y;j++){
                s[j]++;
            }
        }
        int max=0;
        for(int i=0;i<200;i++){
            if(s[i]>max){
                max=s[i];
            }
        }
        printf("%d\n",max*10);
    }
    return 0;
}