HDU_1050 Moving Tables 【贪心】

题目信息：

Moving Tables

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

102030

解题思路：

题目意思是说要讲桌子从一个房间搬到另一个房间，但是因为走廊过窄，仅仅可以容纳一个桌子，给多组数据问最少可以消耗多少时间把桌子都搬完，也就是来判断区间的使用重叠次数。但是有一个细节就是标记数组必须是以走廊为标记及标记数组大小为200而不是以房间为标记数组。（房间1,2都是走廊下标为1）。把测试数据的区间在标记数组上加一，最后求标记数组上大于1的区间个数即可，这里要注意区间如果大于二，需要加上该区间的最大值。

代码如下：

#include <iostream>#include <cstdio>#include <cstring>#include <set>#include <algorithm>#include <map>using namespace std;int change(int n)//将房间换成走廊区间{    if(n%2==0)        return n/2;    else        return(n+1)/2;}int main(){    int vis[205];    int T;    cin>>T;    while(T--)    {        int n,i,j,s,e,t;        cin>>n;        memset(vis,0,sizeof(vis));        for(i=1;i<=n;i++)        {            cin>>s>>e;            if(s>e)            {                t=s;                s=e;                e=t;            }            for(j=change(s);j<=change(e);j++)                vis[j]++;        }        int flag=0,Max=0;        int ans=0;        for(i=1;i<=200;i++)        {            if(vis[i]>1)                flag=1;            if(flag==1&&vis[i]>1)            {                if(vis[i]>Max)                    Max=vis[i];            }            if(vis[i]<=1)            {                flag=0;                if(Max==0)                    continue;                ans+=Max-1;                Max=0;            }        }        cout<<(ans+1)*10<<endl;    }    return 0;}