poj 3347 Kadj Squares(线段切割)

Kadj Squares

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 1911 Accepted: 752

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, and S₄ 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 n integers 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

43 5 1 432 1 20

Sample Output

1 2 41 3

Source

Tehran 2006

题目：http://poj.org/problem?id=3347

题意：一堆正方形，按给定顺序尽量靠左摆放，问最后从上面看有哪几个被看到

分析：首先求出每个正方形的坐标，然后大一点的正方形一定在小一点的正方形上面，把每个正方形转换成一条线段，按正方形的由小到大排序，然后做线段切割即可

代码：

#include<cmath>#include<cstdio>#include<iostream>#include<algorithm>using namespace std;const int mm=55;struct seg{    int id,l,r;}g[mm];int a[mm],b[mm];int i,j,n;bool cmp(seg u,seg v){    return a[u.id]<a[v.id];}bool ok(int i,int l,int r){    if(i>=n)return 1;    if(r<=g[i].l||l>=g[i].r)    {        if(ok(i+1,l,r))return 1;    }    else if(l<g[i].l)    {        if(ok(i+1,l,g[i].l))return 1;    }    else if(r>g[i].r)    {        if(ok(i+1,g[i].r,r))return 1;    }    return 0;}int main(){    while(scanf("%d",&n),n)    {        for(i=0;i<n;++i)        {            scanf("%d",&a[i]);            b[i]=a[i]*2;            for(j=0;j<i;++j)                if(a[i]>=a[j])b[i]=max(b[i],b[j]+a[j]*4);                else b[i]=max(b[i],b[j]+a[i]*4);            g[i].id=i;            g[i].l=b[i]-a[i]*2;            g[i].r=b[i]+a[i]*2;        }        sort(g,g+n,cmp);        for(j=i=0;i<n;++i)            if(ok(i+1,g[i].l,g[i].r))                a[j++]=g[i].id+1;        sort(a,a+j);        for(i=0;i<j;++i)            printf("%d%c",a[i],i<j-1?' ':'\n');    }    return 0;}