HDU 1050 Moving Tables(贪心)

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

10
20
30

 

Answer

将桌子从某个房间搬到另一个房间，搬的时候不能交叉（坑x1，同一房间算，对面房间也算，例如1-2和2-3冲突，1-3和4-5冲突）。另外输入（坑x1），不一定是从小号的房间搬到大号的房间。贪心的思想，起始房间号从小到大排序后，先搬第一张桌子（搬了就erase，因为数据不多），然后往后搬所有不冲突的桌子。直到数组里没有桌子（也就是搬完了）。

 

#include <cstdio>
#include <iostream>
#include <string>
#include <sstream>
#include <cstring>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <map>
#define PI acos(-1.0)
#define ms(a) memset(a,0,sizeof(a))
#define msp memset(mp,0,sizeof(mp))
#define msv memset(vis,0,sizeof(vis))
using namespace std;
//#define LOCAL
struct Node
{
    int b;
    int e;
}t;
int cmp(Node a,Node b)
{
    if(a.b==b.b)return a.e<b.e;
    return a.b<b.b;
}
vector<Node> v;
int main()
{
#ifdef LOCAL
    freopen("in.txt", "r", stdin);
#endif // LOCAL
    ios::sync_with_stdio(false);
    int N;
    cin>>N;
    while(N--)
    {
        int n;
        cin>>n;
        v.clear();
        while(n--)
        {cin>>t.b>>t.e;
        if(t.e<t.b)swap(t.e,t.b);
        v.push_back(t);}
        sort(v.begin(),v.end(),cmp);
        int ans=0;
        while(v.size())
        {
            t=v[0],ans++,v.erase(v.begin());
            for(int i=0;i<(int)v.size();i++)
            {
                if(v[i].b<=t.e)continue;//不好描述，算了
                if((t.e+1)%2==0&&v[i].b-1==t.e)continue;//奇数房的对面房间
                else t.e=v[i].e,v.erase(v.begin()+i),i--;//原来上一句是没有的，WA
            }
        }
        printf("%d\n",ans*10);
    }
    return 0;
}

View Code