贪心2 HDU - 1050 题解

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
题目概要
利用最快的时间完成题目给出的桌子搬运操作，每一次的桌子搬运会占据两个房间之间的走廊十分钟。
题目解析
我们可以先假设自己已经给出了正确的答案，答案的图形表示为：
———— ————   ————
  —— —— ——  ——————
  ————    ————     ————
  ……..
  在上图中每一条横线表示一次搬运操作，在同一行的横线表示一起进行的搬运操作。
  它具有一下性质：
  1.任何一个操作都不能再加入到除本行以外的其它行中。
  2.同一行中的操作没有交点。
  所以我们可以发现只要求出相对应的行数乘以每次搬运的时间10分钟就可以求出答案。而行数为所有操作最多的相交次数。
  结论：求出在同一个地方先交次数的最大值*10即为答案。
代码

#include<iostream>#include<cstdio>#include<vector>using namespace std;int main(){    int T;    scanf("%d",&T);    while(T--){        int N;        scanf("%d",&N);        int array[202]={0};        int max=0;        while(N--){            int left,right;            scanf("%d %d",&left,&right);            left=(left+1)/2;            right=(right+1)/2;            if(left>right){                int temp=right;                right=left;                left=temp;            }            for(int i=left;i<=right;i++){                array[i]++;                if(array[i]>max)max=array[i];            }        }        cout<<max*10<<endl;    }    return 0;}