poj3347 Kadj Squares (计算几何)

D - Kadj Squares

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status

Description

In this problem, you are given a sequence S₁, S₂, ..., S_n 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 b_i be the x coordinates of the bottom vertex of S_i. First, put S₁ such that its left vertex lies on x=0. Then, put S₁, (i > 1) at minimum b_i such that

• b_i > b_i-1 and
• the interior of S_i does not have intersection with the interior of S₁...S_i-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, andS₄ have this property. More formally, S_i is visible from above if it contains a point p, such that no square other than S_i 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 nintegers between 1 to 30, where the ith number is the length of the sides of S_i. 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

题意：变长不同的n个正方形，斜45度按顺序平放在坐标轴上，尽量靠左但不能跃出x=0，

　　　问从上往下看，哪些正方形是可见的。

题解：1、假如前i-1个正方形位置都确定了，那么可以让第i个正方形与前i-1个正方形每个

　　　都计算一次它如果和它相依靠的话左边坐标的值，然后取一个最大的便是这个正方形

　　　的左端点位置。

　　　2、对于j<i的正方形，如果i的边长大于j那么j的最右能看到的部分就不会比i的最左端

　　　点大，反之，i的最左能看到的部分就不会比j最右端点小。

　　　3、通过第2步筛选，将那些最左能看到的端点比最右能看到端点大或等于的去掉，剩

　　　下的就是所要求的。

注意：为避免浮点数运算，根号2约掉了，代码中的边长实际上指的就是边长/根号2，也可以

　　　将该题理解为将正方形投影到x轴上。

#include <iostream>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
struct sa
{
    int l,r,len;
}data[55];
int main()
{
    int n;
    while(cin>>n&&n)
    {
        memset(data,0,sizeof(data));
        for(int i=0;i<n;i++)
        {
            cin>>data[i].len;
            for(int j=0;j<i;j++)
            data[i].l=max(data[i].l,data[j].r-abs(data[i].len-data[j].len));//求左端点，根据等腰直角三角形的性质可推出关系
            data[i].r=data[i].l+2*data[i].len;
        }
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<i;j++)
            {
                if(data[j].r>data[i].l)//可更新
                {
                    if(data[j].len<data[i].len)
                    data[j].r=data[i].l;//右面的把左面的挡住了一块
                    else
                    data[i].l=data[j].r;//左面的把右面的挡住了一块
                }
            }
        }
        for(int i=0;i<n;i++)
        {
            if(data[i].l<data[i].r)
                cout<<i+1<<" ";
        }
        cout<<endl;
    }
    return 0;
}