一.题目
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 25987 | Accepted: 8660 |
Description
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
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 3 + N -rd
line, the remaining test cases are listed in the same manner as above.
Output
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
二.解题技巧
三.实现代码
#include <iostream>
#include <algorithm>
using namespace std;
void InitHist(int* Hist, int len)
{
for (int Index = 0; Index < len; Index++)
{
Hist[Index] = 0;
}
}
int findMax(int* Array, int Len)
{
int Max = 0;
for (int Index = 0; Index < Len; Index++)
{
if (Array[Index] > Max)
{
Max = Array[Index];
}
}
return Max;
}
int main()
{
int NumberOfTestCases = 0;
int Size = 0;
int Hist[200];
int a = 0;
int b = 0;
cin >> NumberOfTestCases;
for (int Index = 0; Index < NumberOfTestCases; Index++)
{
cin >> Size;
InitHist(Hist, 200);
for (int Index = 0; Index < Size; Index++)
{
cin >> a;
cin >> b;
if (b < a)
{
a += b;
b = a - b;
a = a - b;
}
a = (a + 1) / 2;
b = (b + 1) / 2;
s
for (int IndexOfArray = a; IndexOfArray <= b; IndexOfArray++)
{
Hist[IndexOfArray - 1]++;
}
}
cout << findMax(Hist, 200) * 10 << endl;
}
return 0;
}
四.体会
