zoukankan      html  css  js  c++  java
  • Holedox Eating HDU4302 模拟

    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
     
    Recommend
    zhuyuanchen520   |   We have carefully selected several similar problems for you:  4301 4303 4304 4308 4305 
    #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
    ", cas, ans);
        }
    }
  • 相关阅读:
    C++11新特性
    Qt操作xml
    指针和引用的区别
    QT软件主题切换
    常见的临时变量的生成场景
    QQuickWidget+QML设置背景透明
    idea maven Could not transfer artifact
    Java项目启动时执行指定方法的几种方式
    解决bootstrap-table在切换分页后再次查询报错404问题
    bootstrap:表单必填项*标识,及提交前校验
  • 原文地址:https://www.cnblogs.com/joeylee97/p/7289487.html
Copyright © 2011-2022 走看看