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  • 洛谷P2947 [USACO09MAR]仰望Look Up

    P2947 [USACO09MAR]仰望Look Up

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      • 122提交
    • 题目提供者洛谷OnlineJudge
    • 标签USACO2009云端
    • 难度普及/提高-
    • 时空限制1s / 128MB

      讨论  题解  

    最新讨论更多讨论

    • 中文翻译应当为向右看齐
    • 题目中文版范围。。

    题目描述

    Farmer John's N (1 <= N <= 100,000) cows, conveniently numbered 1..N, are once again standing in a row. Cow i has height H_i (1 <= H_i <= 1,000,000).

    Each cow is looking to her left toward those with higher index numbers. We say that cow i 'looks up' to cow j if i < j and H_i < H_j. For each cow i, FJ would like to know the index of the first cow in line looked up to by cow i.

    Note: about 50% of the test data will have N <= 1,000.

    约翰的N(1≤N≤10^5)头奶牛站成一排,奶牛i的身高是Hi(l≤Hi≤1,000,000).现在,每只奶牛都在向左看齐.对于奶牛i,如果奶牛j满足i<j且Hi<Hj,我们可以说奶牛i可以仰望奶牛j. 求出每只奶牛离她最近的仰望对象.

    Input

    输入输出格式

    输入格式:
    • Line 1: A single integer: N

    • Lines 2..N+1: Line i+1 contains the single integer: H_i
    输出格式:
    • Lines 1..N: Line i contains a single integer representing the smallest index of a cow up to which cow i looks. If no such cow exists, print 0.

    输入输出样例

    输入样例#1:
    6 
    3 
    2 
    6 
    1 
    1 
    2 
    
    输出样例#1:
    3 
    3 
    0 
    6 
    6 
    0 
    

    说明

    FJ has six cows of heights 3, 2, 6, 1, 1, and 2.

    Cows 1 and 2 both look up to cow 3; cows 4 and 5 both look up to cow 6; and cows 3 and 6 do not look up to any cow.

    分析:和前几题差不多,不过就是要记录一下每次栈顶的位置.

    #include <cstdio>
    #include <string>
    #include <cstring>
    #include <iostream>
    #include <algorithm>
    
    using namespace std;
    
    int n,h[100010],ans[100010],stk[100010],top,num[100010];
    
    int main()
    {
        scanf("%d", &n);
        for (int i = 1; i <= n; i++)
            scanf("%d", &h[i]);
    
        for (int i = n; i >= 1; i--)
        {
            while (top != 0 && stk[top] <= h[i])
                top--;
            ans[i] = num[top];
            stk[++top] = h[i];
            num[top] = i;
        }
        for (int i = 1; i <= n; i++)
            printf("%d
    ", ans[i]);
    
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/zbtrs/p/7050063.html
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