Moving Tables

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 35371    Accepted Submission(s): 11639


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
Source
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1051 1052 1789 1045 1053  打算從今天開始好好寫部落格（2017-5-20 00;23），

一開始閱讀這個題  感覺有點像並查集，然後仔細想了想，貪心就可以解決。

貪心策略：根據他給的提議，跟往哪移動的方向無關，所以我們把起點弄成相對較小的，對他們進行按起點排序，最後，開始掃描，從第一個開始，把所有能一起發生的一起發生，由於之前排序過，很明顯這樣可以保證最優。注意有個use變數的運用，這個時候貪心能成功的關鍵。

ac程式碼：

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <algorithm>

using namespace std;

struct node{
    int head,tail;
    bool use;
}nn[205];

bool cmp(node x,node y){
    return x.head < y.head;
}

int main()
{
    int t,n;

    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        for(int i=0;i<n;i++){
            scanf("%d%d",&nn[i].head,&nn[i].tail);
        }
        for(int i=0;i<n;i++){
            if(nn[i].head%2)
                nn[i].head++;
            if(nn[i].tail%2)
                nn[i].tail++;

            if(nn[i].head > nn[i].tail){
                int tmp;
                tmp=nn[i].head;
                nn[i].head=nn[i].tail;
                nn[i].tail=tmp;
            }

            nn[i].use=true;
        }
        sort(nn,nn+n,cmp);
        int ans=0;
        for(int i=0;i<n ;i++){
            int tmp=nn[i].tail;

            if(nn[i].use == false)
                continue ;

            for(int j=i+1;j<n;j++){
                if(nn[j].use==false)
                    continue ;

                if(nn[j].head > tmp){
                    nn[j].use=false;
                    tmp=nn[j].tail;
                }
            }

            ans+=10;
        }

        printf("%d\n",ans);
    }


    return 0;
}