POJ 3347 Kadj Squares

2019年1月25日 323次阅读 来源: kuangbin

Kadj Squares

Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 2132 Accepted: 843

Description

In this problem, you are given a sequence S1, S2, …, Sn of squares of different sizes. The sides of the squares are integer numbers. We locate the squares on the positive x–y quarter of the plane, such that their sides make 45 degrees with x and y axes, and one of their vertices are on y=0 line. Let bi be the x coordinates of the bottom vertex of Si. First, put S1 such that its left vertex lies on x=0. Then, put S1, (i > 1) at minimum bi such that

  • bi > bi-1 and
  • the interior of Si does not have intersection with the interior of S1…Si-1.

 

 

The goal is to find which squares are visible, either entirely or partially, when viewed from above. In the example above, the squares S1, S2, and S4 have this property. More formally, Si is visible from above if it contains a point p, such that no square other than Si intersect the vertical half-line drawn from p upwards.

Input

The input consists of multiple test cases. The first line of each test case is n (1 ≤ n ≤ 50), the number of squares. The second line contains n integers between 1 to 30, where the ith number is the length of the sides of Si. The input is terminated by a line containing a zero number.

Output

For each test case, output a single line containing the index of the visible squares in the input sequence, in ascending order, separated by blank characters.

Sample Input

4
3 5 1 4
3
2 1 2
0

Sample Output

1 2 4
1 3

Source

Tehran 2006           这题是在做计算几何的时候遇到的。   但是发现有简单的做法，不需要用计算几何。   为了避免浮点数运算，把长度乘以sqrt(2).   具体看代码吧，代码看得很清楚了。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;

const int MAXN = 55;
struct Node
{
    int l,r,len;
};
Node node[MAXN];
int main()
{
    int n;
    while(scanf("%d",&n) == 1 && n)
    {
        for(int i = 1;i <= n;i++)
        {
            scanf("%d",&node[i].len);
            node[i].l = 0;
            for(int j = 1;j < i;j++)
                node[i].l = max(node[i].l,node[j].r - abs(node[i].len - node[j].len));
            node[i].r = node[i].l + 2*node[i].len;
        }
        for(int i = 1;i <= n;i++)
        {
            for(int j = 1;j < i;j++)
                if(node[i].l < node[j].r && node[i].len < node[j].len)
                     node[i].l = node[j].r;
            for(int j = i+1;j <= n;j++)
                if(node[i].r > node[j].l && node[i].len < node[j].len)
                     node[i].r = node[j].l;
        }
        bool first = true;
        for(int i = 1;i <= n;i++)
            if(node[i].l < node[i].r)
        {
            if(first)first = false;
            else printf(" ");
            printf("%d",i);
        }
        printf("\n");
    }
    return 0;
}

 

     

    原文作者：kuangbin
    原文地址: https://www.cnblogs.com/kuangbin/p/3192475.html
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