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描述

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.

输入

The input consists of T test cases. The number of test cases ) (T is given in the first line of the input file. 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 3 + N -rd

line, the remaining test cases are listed in the same manner as above.

输出

The output should contain the minimum time in minutes to complete the moving, one per line.

样例输入

3 4 10 20 30 40 50 60 70 80 2 1 3 2 200 3 10 100 20 80 30 50

样例输出

1020

30

思路1：

把房间编号转换为直线上的线段，那么该问题就转换为了区间问题，类比 多个活动至少需要几个房间的问题

可以按照区间起始位置升序排列，然后用优先队列记录结束时间

#include
     
      using namespace std;struct Node{	int s,e;	bool operator<(const Node o)const{		return s < o.s;	}}nodes[200+5];priority_queue
      
       
        ,greater
        
          > pq;int main()	{		int t,n;		scanf("%d",&t);		while(t--){			scanf("%d",&n);			int s,e;			for(int i = 0; i < n; ++i){				scanf("%d%d",&s,&e);				if(s > e){					int t = s;					s = e;					e = t;				}				nodes[i].s = s;				nodes[i].e = e;				if(s%2 == 0)					nodes[i].s = s-1;//看图就知道了 区间转化				if(e%2 == 1)					nodes[i].e = e+1;			}			sort(nodes,nodes+n);			while(!pq.empty())//注意每次需要清空队列				pq.pop(); 			int ans = 0;			for(int i = 0; i < n; ++i){				if(!pq.empty()){					int end = pq.top();					if(nodes[i].s >= end){						pq.pop();						pq.push(nodes[i].e);					}					else{						++ans;						pq.push(nodes[i].e);					}				}				else{					++ans;					pq.push(nodes[i].e);				}			}			printf("%d\n",ans*10);					}						return 0;	}
        
       
      
     

思路二：找到最大重叠的部分 就是

#include
     
        #include
      
         #include
       
          using namespace std;  int main(){      int t,n,count[410],i,start,end,k;      scanf("%d",&t);      while(t--){          scanf("%d",&n);          memset(count,0,sizeof(count));          while(n--){              scanf("%d%d",&start,&end);              //可能出发位置比目的地房间大。              if(start>end){                  //无论大小，我们都可以看做从小的房间移动到大的房间                  k=start;                  start=end;                  end=k;              }              if(start%2==0)//考虑实际情况，出发房间为偶数是减一，可参照题中给出的图一                 start-=1;              if(end%2==1)//目的地房间为奇数时加一                 end+=1;              for(i=start;i<=end;++i)                 count[i]+=10;          }          printf("%d\n",*max_element(count,count+400));//STL中寻找数列最大值函数      }      return 0;  }
       
      
     

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