计蒜客 Zoning Houses(线段树区间最大次大)

Given a registry of all houses in your state or province, you would like to know the minimum size of an axis-aligned square zone such that every house in a range of addresses lies in the zone or on its border. The zoning is a bit lenient and you can ignore any one house from the range to make the zone smaller.

The addresses are given as integers from 1..n. Zoning requests are given as a consecutive range of houses. A valid zone is the smallest axis-aligned square that contains all of the points in the range, ignoring at most one.

Given the (x, y) locations of houses in your state or province, and a list of zoning requests, you must figure out for each request: What is the length of a side of the smallest axis-aligned square zone that contains all of the houses in the zoning request, possibly ignoring one house?

Input Format

Each input will consist of a single test case.

Note that your program may be run multiple times on different inputs.

Each test case will begin with a line containing two integers nn and q(1≤n,q≤10^5), where n is the number of houses, and q is the number of zoning requests.

The next n lines will each contain two integers, x and y(−10^9≤x,y≤10^9), which are the (x,y) coordinates of a house in your state or province. The address of this house corresponds with the order in the input. The first house has address 1, the second house has address 2, and so on. No two houses will be at the same location.

The next q lines will contain two integers a and b(1≤a<b≤n), which represents a zoning request for houses with addresses in the range [a..b] inclusive.

Output Format

Output q lines.

On each line print the answer to one of the zoning requests, in order: the side length of the smallest axis-aligned square that contains all of the points of houses with those addresses, if at most one house can be ignored.

样例输入1

样例输出1

1

0

样例输入2

样例输出2

300

299

题意

n个点q次询问，每次询问编号[a,b]的点中，最多忽略一个点覆盖剩下点的最小正方形。

题解

如果不忽略点，那么就是max(最大的X-最小的X，最大的Y-最小的Y)。

如果忽略点，那么就有4种情况，最大的X，最大的Y，最小的X，最小的Y，那么就是次大，次小。

还有4种情况，忽略的点同时是最大的X最大的Y，最大的X最小的Y，最小的X最小的Y，最小的X最大的Y。

线段树维护比较复杂，要自己仔细推一下。

代码

 #include<bits/stdc++.h>

 using namespace std;

 const int N=1e5+;

 int x[N],y[N];

 map<pair<int,int>,int>ma;

 int x_maxx[N<<],x_maxx1[N<<],x_minn[N<<],x_minn1[N<<];

 int y_maxx[N<<],y_maxx1[N<<],y_minn[N<<],y_minn1[N<<];

 void up(int rt)

 {

     int ls=rt<<,rs=rt<<|;

     x_maxx1[rt]=max(min(x_maxx[ls],x_maxx[rs]),max(x_maxx1[ls],x_maxx1[rs]));

     x_maxx[rt]=max(x_maxx[ls],x_maxx[rs]);

     x_minn1[rt]=min(max(x_minn[ls],x_minn[rs]),min(x_minn1[ls],x_minn1[rs]));

     x_minn[rt]=min(x_minn[ls],x_minn[rs]);

     y_maxx1[rt]=max(min(y_maxx[ls],y_maxx[rs]),max(y_maxx1[ls],y_maxx1[rs]));

     y_maxx[rt]=max(y_maxx[ls],y_maxx[rs]);

     y_minn1[rt]=min(max(y_minn[ls],y_minn[rs]),min(y_minn1[ls],y_minn1[rs]));

     y_minn[rt]=min(y_minn[ls],y_minn[rs]);

 }

 int X_MAX,X_MAX1,X_MIN,X_MIN1;

 int Y_MAX,Y_MAX1,Y_MIN,Y_MIN1;

 void query(int L,int R,int l,int r,int rt)

 {

     if(L<=l&&r<=R)

     {

         //printf("FUCK=%d\n%d %d\n",rt,X_MIN,X_MIN1);

         X_MAX1=max(min(X_MAX,x_maxx[rt]),max(X_MAX1,x_maxx1[rt]));

         X_MAX=max(X_MAX,x_maxx[rt]);

         X_MIN1=min(max(X_MIN,x_minn[rt]),min(X_MIN1,x_minn1[rt]));

         X_MIN=min(X_MIN,x_minn[rt]);

         Y_MAX1=max(min(Y_MAX,y_maxx[rt]),max(Y_MAX1,y_maxx1[rt]));

         Y_MAX=max(Y_MAX,y_maxx[rt]);

         Y_MIN1=min(max(Y_MIN,y_minn[rt]),min(Y_MIN1,y_minn1[rt]));

         Y_MIN=min(Y_MIN,y_minn[rt]);

         //printf("rt=%d\n%d %d\n",rt,X_MIN,X_MIN1);

         return;

     }

     int mid=(l+r)>>;

     if(L<=mid)query(L,R,l,mid,rt<<);

     if(R>mid)query(L,R,mid+,r,rt<<|);

 }

 void build(int l,int r,int rt)

 {

     if(l==r)

     {

         x_maxx[rt]=x_minn[rt]=x[l];

         x_maxx1[rt]=-2e9;

         x_minn1[rt]=2e9;

         y_maxx[rt]=y_minn[rt]=y[l];

         y_maxx1[rt]=-2e9;

         y_minn1[rt]=2e9;

         return;

     }

     int mid=(l+r)>>;

     build(l,mid,rt<<);

     build(mid+,r,rt<<|);

     up(rt);

 }

 int main()

 {

     int n,q;

     scanf("%d%d",&n,&q);

     for(int i=;i<=n;i++)

     {

         scanf("%d%d",&x[i],&y[i]);

         ma[{x[i],y[i]}]=i;

     }

     build(,n,);

     /*

     for(int i=1;i<=7;i++)

         printf("i=%d\n%d %d %d %d\n%d %d %d %d\n",i,x_maxx[i],x_maxx1[i],x_minn[i],x_minn1[i],

         y_maxx[i],y_maxx1[i],y_minn[i],y_minn1[i]);

     */

     while(q--)

     {

         int l,r;

         scanf("%d%d",&l,&r);

         X_MAX=X_MAX1=Y_MAX=Y_MAX1=-2e9;

         X_MIN=X_MIN1=Y_MIN=Y_MIN1=2e9;

         query(l,r,,n,);

         //printf("%d %d %d %d\n",X_MAX,X_MAX1,X_MIN,X_MIN1);

         //printf("%d %d %d %d\n",Y_MAX,Y_MAX1,Y_MIN,Y_MIN1);

         int ans=max(X_MAX-X_MIN,Y_MAX-Y_MIN);

         ans=min(ans,max(X_MAX1-X_MIN,Y_MAX-Y_MIN));

         ans=min(ans,max(X_MAX-X_MIN1,Y_MAX-Y_MIN));

         ans=min(ans,max(X_MAX-X_MIN,Y_MAX1-Y_MIN));

         ans=min(ans,max(X_MAX-X_MIN,Y_MAX-Y_MIN1));

         if(l<=ma[{X_MAX,Y_MAX}]&&ma[{X_MAX,Y_MAX}]<=r)ans=min(ans,max(X_MAX1-X_MIN,Y_MAX1-Y_MIN));

         if(l<=ma[{X_MAX,Y_MIN}]&&ma[{X_MAX,Y_MIN}]<=r)ans=min(ans,max(X_MAX1-X_MIN,Y_MAX-Y_MIN1));

         if(l<=ma[{X_MIN,Y_MAX}]&&ma[{X_MIN,Y_MAX}]<=r)ans=min(ans,max(X_MAX-X_MIN1,Y_MAX1-Y_MIN));

         if(l<=ma[{X_MIN,Y_MIN}]&&ma[{X_MIN,Y_MIN}]<=r)ans=min(ans,max(X_MAX-X_MIN1,Y_MAX-Y_MIN1));

         printf("%d\n",ans);

     }

     return ;

 }

 /*

 5 5

 -1000000000 -1000000000

 -1000000000 1000000000

 1000000000 1000000000

 1000000000 -1000000000

 0 0

 1 3

 */