先把矩形擴(kuò)大 sqrt(2) 倍，轉(zhuǎn)化為整點(diǎn)問題。然后逐個(gè)求出每個(gè)矩形的坐標(biāo)。
對(duì)于每個(gè)矩形分別求出在它之上的矩形覆蓋的區(qū)間大小 t1，和包括它本身以及在它之上的矩形覆蓋的區(qū)間大小 t2
若 t1 == t2，則該矩形被遮蓋。
/**//*************************************************************************
Author: WHU_GCC
Created Time: 2007-8-15 20:48:05
File Name: pku3347.cpp
Description:
************************************************************************/
#include <iostream>
using namespace std;
#define out(x) (cout << #x << ": " << x << endl)
typedef long long int64;
const int maxint = 0x7FFFFFFF;
const int64 maxint64 = 0x7FFFFFFFFFFFFFFFLL;
template <class T> void show(T a, int n) 
{ for (int i = 0; i < n; ++i) cout << a[i] << ' '; cout << endl; }
template <class T> void show(T a, int r, int l) { for (int i = 0; i < r; ++i) show(a[i], l); cout << endl; }
typedef struct square_t
{
int c, r;
};
typedef struct interval_t
{
int l, r;
};
bool operator <(const interval_t &a, const interval_t &b)
{
return a.l < b.l || a.l == b.l && a.r < b.r;
}
square_t s[100];
interval_t interval[100];
int cnt_interval;
int merge()
{
if (cnt_interval == 0) return 0;
sort(interval, interval + cnt_interval);
int l = interval[0].l, r = interval[0].r;
int len = 0;
for (int i = 1; i < cnt_interval; i++)
if (interval[i].l <= r)
r >?= interval[i].r;
else
{
len += r - l;
l = interval[i].l;
r = interval[i].r;
}
len += r - l;
return len;
}
int ans[100];
int cnt_ans;
int main()
{
int n;
while (scanf("%d", &n), n != 0)
{
for (int i = 0; i < n; i++)
scanf("%d", &s[i].r);
s[0].c = s[0].r;
for (int i = 1; i < n; i++)
{
int t = 0;
for (int j = 0; j < i; j++)
t >?= min(s[j].c + 2 * s[j].r, s[j].c + 2 * s[i].r);
s[i].c = t;
}
cnt_ans = 0;
for (int i = 0; i < n; i++)
{
cnt_interval = 0;
for (int j = 0; j < n; j++) if (s[j].r > s[i].r)
{
interval[cnt_interval].l = s[j].c - s[j].r;
interval[cnt_interval].r = s[j].c + s[j].r;
cnt_interval++;
}
int t1 = merge();
cnt_interval = 0;
for (int j = 0; j < n; j++) if (s[j].r > s[i].r || j == i)
{
interval[cnt_interval].l = s[j].c - s[j].r;
interval[cnt_interval].r = s[j].c + s[j].r;
cnt_interval++;
}
int t2 = merge();
if (t1 != t2)
ans[cnt_ans++] = i;
}
for (int i = 0; i < cnt_ans; i++)
printf("%d ", ans[i] + 1);
printf("\n");
}
return 0;
}