Holedox Eating HDU4302 模拟

转载于 2017-08-05 10:49:00 发布 · 76 阅读

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原文链接：http://www.cnblogs.com/joeylee97/p/7289487.html

本文探讨了一种名为Holedox的小动物在一个管道中寻找并吃掉蛋糕的算法问题。Holedox从起点出发，根据最近的蛋糕位置和上一次移动的方向选择下一个目标，直到所有蛋糕被吃完。文章通过具体样例输入输出，展示了如何计算Holedox移动的总距离。

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

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#include<iostream>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<string>
#include<map>
#include<cstring>
using namespace std;
#define MAXN 500010
#define INF 0x3f3f3f3f
typedef long long LL;
/*
模拟 map每次lower_bound
*/
map<int, int> m;
int p, T, L, n, eat, tmp, ans;
int main()
{
    scanf("%d", &T);
    for(int cas = 1;cas<=T;cas++)
    {
        m.clear();
        scanf("%d%d", &L, &n);
        ans = 0, p = 0;
        int add1 ,add2;
        bool dir;
        for (int i = 0; i < n; i++)
        {
            scanf("%d", &eat);
            if (eat)
            {
                if (m.empty()) continue;
                add1 = add2 = INF;
                map<int, int>::iterator it = m.lower_bound(p);
                if (it != m.end())
                    add1 = min(add1, abs(it->first - p));
                if (it != m.begin())
                    it--, add2 = min(add2, abs(it->first - p));
                if (add1 != INF&&add1 == add2)
                {
                    if (dir) p = p + add1;
                    else p = p - add1;
                    ans += add1;
                }
                else if (add1 < add2)
                {
                    if (add1) dir = true;
                    p = p + add1, ans += add1;
                }
                else
                {
                    if (add2) dir = false;
                    p = p - add2, ans += add2;
                }
                if (--m[p] == 0)
                    m.erase(p);
            //    cout << ':' << ans <<"at" << p << endl;
            }
            else
                scanf("%d", &tmp), m[tmp]++;
        }
        printf("Case %d: %d\n", cas, ans);
    }
}