POJ 3263-Tallest Cow

最新推荐文章于 2021-10-06 20:34:55 发布

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本文探讨了一道经典的算法题目，即根据特定条件确定每头奶牛的最大可能身高。题目中给出了奶牛的数量、最高峰的高度及位置以及一系列视线约束条件。通过逐步调整奶牛身高来确保所有已知条件成立。

Description

FJ’s N (1 ≤ N ≤ 10,000) cows conveniently indexed 1..N are standing in a line. Each cow has a positive integer height (which is a bit of secret). You are told only the height H (1 ≤ H ≤ 1,000,000) of the tallest cow along with the index I of that cow.

FJ has made a list of R (0 ≤ R ≤ 10,000) lines of the form “cow 17 sees cow 34”. This means that cow 34 is at least as tall as cow 17, and that every cow between 17 and 34 has a height that is strictly smaller than that of cow 17.

For each cow from 1..N, determine its maximum possible height, such that all of the information given is still correct. It is guaranteed that it is possible to satisfy all the constraints.

Input

Line 1: Four space-separated integers: N, I, H and R
Lines 2..R+1: Two distinct space-separated integers A and B (1 ≤ A, B ≤ N), indicating that cow A can see cow B.
Output

Lines 1..N: Line i contains the maximum possible height of cow i.
Sample Input

9 3 5 5
1 3
5 3
4 3
3 7
9 8
Sample Output

5
4
5
3
4
4
5
5
5
Source

USACO 2007 January Silver
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分析

一开始全部高度是最高高度，每输入a看见b，那么如果a和b之间所有奶牛有一头比a高，那么a与b之间所有奶牛全部减1。
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程序：

#include<iostream>
using namespace std;
int main()
{
    int n,tst,h,r,s[10001],x,y,p,k;
    cin>>n>>tst>>h>>r;
    for (int i=1;i<=n;i++)
    s[i]=h;
    for (int i=1;i<=r;i++)
    {
        cin>>x>>y;
        p=x;
        if (x>y)
        {
            k=x;x=y;y=k;
        }
        for (int j=x+1;j<=y-1;j++)
        if (s[j]>=s[p])
        {
            for (int k=x+1;k<=y-1;k++)
            s[k]--;
            break;
        }
    }
    for (int i=1;i<=n;i++)
    cout<<s[i]<<endl;
    return 0;
}