链接

这题其实与几何没太大关系,还不错的题目。

参考吴永辉的算法设计书。

用lefi、rigi分别表示正方形在x轴上的投影。

为了避免用小数,把边长都扩大sqrt(2)倍,这样lef1 = 0,rig1 = 2*a1;

lefi = max{rigj-abs(ai-aj)}

rigi = lefi+2*ai;

求出各个正方形的投影之后,这题就好做了。用le和re表示正方形的可见区间。

le = max(rigj,lefi) j <i

re = min(lefj,rigi) j>i

 #include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<cmath>
#include<queue>
#include<set>
using namespace std;
#define N 55
#define LL long long
#define INF 0xfffffff
const double eps = 1e-;
const double pi = acos(-1.0);
const double inf = ~0u>>;
int o[N],a[N],lef[N],rig[N]; int main()
{
int i,j,n;
while(scanf("%d",&n)&&n)
{
memset(o,,sizeof(o));
for(i = ; i <= n ;i++)
{
scanf("%d",&a[i]);
}
lef[] = ;
rig[] = *a[];
for(i = ; i <= n ;i++)
{
lef[i] = ;
for(j = ; j < i ; j++)
lef[i] = max(lef[i],rig[j]-abs(a[i]-a[j]));
rig[i] = lef[i]+*a[i];
//cout<<lef[i]<<" "<<rig[i]<<endl;
}
int g = ;
for(i = ; i <= n ;i++)
{
int le = lef[i],re = rig[i];
for(j = ; j < i ; j++)
{
le = max(le,rig[j]);
}
for(j = i+ ; j <= n; j++)
re = min(re,lef[j]);
//cout<<le<<" "<<re<<endl;
if(re>le)
o[++g] = i;
}
for(i = ; i < g ; i++)
printf("%d ",o[i]);
printf("%d\n",o[g]);
}
return ;
}

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