传送门
题意
在二维平面上有n个点,初始时使用一个矩形将所有点都框起来;现在想用两个矩形将这些点框起来,询问用两个矩形框起来之后的面积相比于初始时一个大矩形面积减少了多少;
题解
因为要用两个矩形框住,那么思考一下就知道这两个矩形只能是左右两个或者是上下两个。
①对于左右两个矩形的情况:
就先将这些点按x值从小到大排序,若x值相同就按照y值从小到大排序;之后枚举以哪个点作为分为左右两部分的分界点,在这两边分别求响应的矩形面积大小。
②对于上下两个矩形的情况:
实际上将x,y值调换一下位置,再根据第一种情况一样的做法就可以了。
那么问题就是怎么找出每个矩形的边界呢,这时候就要请上线段树了(实际好像不用也可以!),开两颗线段树,一颗维护x值的最大最小值,一颗维护y值的最大最小值;找到边界之后,面积就呼之欲出啦!
AC Code(358ms)线段树又臭又长又慢
#include<iostream>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<stack>
#include<cstdio>
#include<string>
#include<algorithm>
using namespace std
;
#define show(x) std :: cerr << #x << " = " << x << std :: endl
#define MAX MAXN
typedef long long ll
;
const int MAXN
= 5e4 + 50;
const ll mo
= 998244353;
const int inf
= 1e9 + 7;
inline int lson(int p
){return p
<< 1;}
inline int rson(int p
){return p
<< 1 | 1;}
struct node
{
int x
,y
;
}p
[MAXN
];
int n
,a
[MAXN
],b
[MAXN
];
ll ans
= 1e18;
int cmp(node a
,node b
)
{
if(a
.x
== b
.x
)return a
.y
< b
.y
;
return a
.x
< b
.x
;
}
struct tree
{
int l
,r
;
int Max
,Min
;
}t1
[MAXN
<< 2],t2
[MAXN
<< 2];
inline push_up1(int p
)
{
t1
[p
].Max
= max(t1
[lson(p
)].Max
,t1
[rson(p
)].Max
);
t1
[p
].Min
= min(t1
[lson(p
)].Min
,t1
[rson(p
)].Min
);
}
inline push_up2(int p
)
{
t2
[p
].Max
= max(t2
[lson(p
)].Max
,t2
[rson(p
)].Max
);
t2
[p
].Min
= min(t2
[lson(p
)].Min
,t2
[rson(p
)].Min
);
}
void build1(int p
,int l
,int r
)
{
t1
[p
].l
= l
,t1
[p
].r
= r
;
if(l
== r
)
{
t1
[p
].Max
= a
[l
];
t1
[p
].Min
= a
[l
];
return;
}
int mid
= l
+ r
>> 1;
build1(lson(p
),l
,mid
);
build1(rson(p
),mid
+ 1,r
);
push_up1(p
);
}
int query_Max1(int p
,int l
,int r
)
{
if(l
<= t1
[p
].l
&& r
>= t1
[p
].r
)
{
return t1
[p
].Max
;
}
int mid
= t1
[p
].l
+ t1
[p
].r
>> 1;
int maxL
= -inf
,maxR
= -inf
;
if(l
<= mid
)maxL
= max(maxL
,query_Max1(lson(p
),l
,r
));
if(r
> mid
)maxR
= max(maxR
,query_Max1(rson(p
),l
,r
));
return max(maxL
,maxR
);
}
int query_Min1(int p
,int l
,int r
)
{
if(l
<= t1
[p
].l
&& r
>= t1
[p
].r
)
{
return t1
[p
].Min
;
}
int mid
= t1
[p
].l
+ t1
[p
].r
>> 1;
int minL
= inf
,minR
= inf
;
if(l
<= mid
)minL
= min(minL
,query_Min1(lson(p
),l
,r
));
if(r
> mid
)minR
= min(minR
,query_Min1(rson(p
),l
,r
));
return min(minL
,minR
);
}
void build2(int p
,int l
,int r
)
{
t2
[p
].l
= l
,t2
[p
].r
= r
;
if(l
== r
)
{
t2
[p
].Max
= b
[l
];
t2
[p
].Min
= b
[l
];
return;
}
int mid
= l
+ r
>> 1;
build2(lson(p
),l
,mid
);
build2(rson(p
),mid
+ 1,r
);
push_up2(p
);
}
int query_Max2(int p
,int l
,int r
)
{
if(l
<= t2
[p
].l
&& r
>= t2
[p
].r
)
{
return t2
[p
].Max
;
}
int mid
= t2
[p
].l
+ t2
[p
].r
>> 1;
int maxL
= -inf
,maxR
= -inf
;
if(l
<= mid
)maxL
= max(maxL
,query_Max2(lson(p
),l
,r
));
if(r
> mid
)maxR
= max(maxR
,query_Max2(rson(p
),l
,r
));
return max(maxL
,maxR
);
}
int query_Min2(int p
,int l
,int r
)
{
if(l
<= t2
[p
].l
&& r
>= t2
[p
].r
)
{
return t2
[p
].Min
;
}
int mid
= t2
[p
].l
+ t2
[p
].r
>> 1;
int minL
= inf
,minR
= inf
;
if(l
<= mid
)minL
= min(minL
,query_Min2(lson(p
),l
,r
));
if(r
> mid
)minR
= min(minR
,query_Min2(rson(p
),l
,r
));
return min(minL
,minR
);
}
void solve()
{
sort(p
+ 1,p
+ 1 + n
,cmp
);
for(int i
= 1;i
<= n
;i
++)
{
a
[i
] = p
[i
].x
;
b
[i
] = p
[i
].y
;
}
build1(1,1,n
);
build2(1,1,n
);
for(int i
= 1;i
<= n
- 1;i
++)
{
int Max_x_i
= query_Max1(1,1,i
);
int Min_x_i
= query_Min1(1,1,i
);
int Max_y_i
= query_Max2(1,1,i
);
int Min_y_i
= query_Min2(1,1,i
);
int Max_x_n
= query_Max1(1,i
+ 1,n
);
int Min_x_n
= query_Min1(1,i
+ 1,n
);
int Max_y_n
= query_Max2(1,i
+ 1,n
);
int Min_y_n
= query_Min2(1,i
+ 1,n
);
ll res
= 1LL * (Max_x_i
- Min_x_i
) * (Max_y_i
- Min_y_i
) + 1LL * (Max_x_n
- Min_x_n
) * (Max_y_n
- Min_y_n
);
ans
= min(ans
,res
);
}
}
int main()
{
scanf("%d",&n
);
int Max_Y
= 0,Min_Y
= inf
;
int Max_X
= 0,Min_X
= inf
;
for(int i
= 1;i
<= n
;i
++)
{
scanf("%d%d",&p
[i
].x
,&p
[i
].y
);
Max_X
= max(Max_X
,p
[i
].x
);
Max_Y
= max(Max_Y
,p
[i
].y
);
Min_X
= min(Min_X
,p
[i
].x
);
Min_Y
= min(Min_Y
,p
[i
].y
);
}
ll ff
= 1LL *(Max_X
- Min_X
) * (Max_Y
- Min_Y
);
solve();
for(int i
= 1;i
<= n
;i
++)swap(p
[i
].x
,p
[i
].y
);
solve();
printf("%lld",ff
- ans
);
}
