MUTC 1 B Holedox Eating STL

Holedox Eating

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2314    Accepted Submission(s): 780

Problem Description

Holedox is a small animal which can be considered as one point. It lives in a straight pipe whose length is L. Holedox can only move along the pipe. Cakes may appear anywhere in the pipe, from time to time. When Holedox wants to eat cakes, it always goes to the nearest one and eats it. If there are many pieces of cake in different directions Holedox can choose, Holedox will choose one in the direction which is the direction of its last movement. If there are no cakes present, Holedox just stays where it is.

 

Input

The input consists of several test cases. The first line of the input contains a single integer T (1 <= T <= 10), the number of test cases, followed by the input data for each test case.The first line of each case contains two integers L,n(1<=L,n<=100000), representing the length of the pipe, and the number of events. 
The next n lines, each line describes an event. 0 x(0<=x<=L, x is a integer) represents a piece of cake appears in the x position; 1 represent Holedox wants to eat a cake.
In each case, Holedox always starts off at the position 0.

 

Output

Output the total distance Holedox will move. Holedox don’t need to return to the position 0.

 

Sample Input

3
10 8
0 1
0 5
1
0 2
0 0
1
1
1 

10 7
0 1
0 5
1
0 2
0 0
1
1

10 8
0 1
0 1
0 5
1
0 2
0 0
1
1

 

Sample Output

Case 1: 9
Case 2: 4
Case 3: 2

 

Author

BUPT

 

Source

2012 Multi-University Training Contest 1

 

Recommend

zhuyuanchen520

 

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STL map中的元素按照从小到大的顺序排序，用迭代器it++和it--到达相邻元素

-----------------

#include <iostream>
#include <cstdio>
#include <map>
#include <algorithm>
#include <cstring>

using namespace std;

const int OO=1e9;

map<int,int>mp;
map<int,int>::iterator it,itr,itl;
map<int,int>::iterator p;
int main()
{
    //freopen("input.txt","r",stdin);
    int T,len,m,tp,x;
    int dir;
    int ans;
    int cas=1;
    int ke;
    scanf("%d",&T);
    while (T--)
    {
        ans=0;
        ke=0;
        mp.clear();
        scanf("%d%d",&len,&m);
        dir=1;
        mp[OO]=1;
        mp[-OO]=1;
        mp[0]=0;
        p=mp.find(0);
        while (m--)
        {
            scanf("%d",&tp);
            if (tp==0)
            {
                scanf("%d",&x);
                mp[x]++;
                ke++;
                continue;
            }
            if (ke==0) continue;
            ke--;
            itl=itr=it=p;
            itl--;
            itr++;
            if (p->second>0)
            {
                p->second--;
                continue;
            }
            else if ( p->first - itl->first < itr->first - p->first &&
                      itl->first!=OO &&
                      itl->first!=-OO )
            {
                ans+=p->first - itl->first;
                p=itl;
                p->second--;
                dir=-1;
            }
            else if ( p->first - itl->first > itr->first - p->first &&
                      itr->first!=OO &&
                      itr->first!=-OO )
            {
                ans+=itr->first - p->first;
                p=itr;
                p->second--;
                dir=1;
            }
            else if ( p->first - itl->first == itr->first - p->first &&
                      itl->first!=OO &&
                      itl->first!=-OO &&
                      itr->first!=OO &&
                      itr->first!=-OO )
            {
                if (dir==1)
                {
                    ans+=itr->first - p->first;
                    p=itr;
                    p->second--;
                }
                else if (dir==-1)
                {
                    ans+=p->first - itl->first;
                    p=itl;
                    p->second--;
                }
            }
            if (it->second==0)
            {
                mp.erase(it);
            }
        }
        printf("Case %d: %d\n",cas++,ans);
    }
    return 0;
}