POJ 1083 && HDU 1050 Moving Tables （贪心）

Moving Tables

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20302    Accepted Submission(s): 6889

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 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

 

Source

Asia 2001, Taejon (South Korea)

解题思路：贪心。考虑每次的最优解，则对面的两房间共有它们所相应的区域，每次交错，都会产生一次冲突，就代表他们不能在同一次搬运，考虑到多次转移的交错的区域，最大交错的次数就代表了最小的搬运次数。

AC代码：

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
using namespace std;

int room[205];

int main(){
//    freopen("in.txt", "r", stdin);
    int t, n, x, y;
    scanf("%d", &t);
    while(t--){
        memset(room, 0, sizeof(room));
        scanf("%d", &n);
        for(int i=0; i<n; i++){
            scanf("%d%d", &x, &y);
            if(x > y) swap(x, y);          //防止起始房间大于终止房间
            for(int i=(x+1)/2; i<=(y+1)/2; i++){
                room[i] ++;                //相应区域的交错次数添加
            }
        }
        int ans = 0;
        for(int i=1; i<=201; i++){
            if(ans < room[i])  ans = room[i];
        }
        printf("%d
", ans*10);
    }
    return 0;
}