HDU 1050 Moving Tables(贪心)

本文介绍了一个针对特定场景下的表格移动问题的解决方案。该方案通过贪心算法实现,旨在减少在狭窄走廊中移动大量表格所需的时间。通过对每个移动操作进行排序并采用贪心策略,确保了在不冲突的情况下同时进行多个移动任务,从而达到最小化总移动时间的目的。

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The famous ACM (Advanced Computer Maker) Company has rented a floor of a building whose shape is in the following figure. 

这里写图片描述

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

这题要注意一下走廊的位置,2n-1和2n是用同一段走廊的,所以输入一个区间时要把它扩大一下,扩大成所占走廊的最小编号和最大编号再贪心做就好了

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
#include <stack>
#include <vector>
#define maxn 10010
#define maxe 100010
typedef long long ll;
using namespace std;
const double eps=1e-5;
const int inf=0x3f3f3f3f3f;
struct Node
{
    int s,t;
    bool done;
    Node(){};
    Node(int a,int b)
    {
        if(a>b)swap(a,b);
        if(b%2==1)b++;
        if(a%2==0)a--;
        s=a;t=b;
        done=0;
    }
    bool operator <(Node ano)
    {
        return s<ano.s;
    }

}node[maxn];
int main()
{
    int n;
    int T;
    int a,b;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&a,&b);
            Node t(a,b);
            node[i]=t;
        }
        sort(node,node+n);
        int lef=n;
        int last=0;
        int ans=0;
        while(lef)
        {
            last=-1;
            for(int i=0;i<n;i++)
            {
                if(node[i].done)continue;
                if(last==-1||node[i].s>node[last].t)
                {
                    node[i].done=true;
                    last=i;
                    lef--;
                }
            }
            ans+=10;
        }
        printf("%d\n",ans);
    }
    return 0;
}
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