CodeForces 846D. Monitor(二维线段树/二维RMQ)

题目大意：

给予n,m,k,t 和 t 行i,j,val代表第i行j列的元素会在val天后损坏。

求在整张（n*m）图中求一个k*k的区域的数全部损坏的最小天数。



试了一下二维线段树，总结一下就是和一维的差别不大，从二叉树变成了四叉树（三维线段树岂不是要八叉树，丧心病狂）。

总体上就是把一个长方形均分四块，然后总的节点数就是一个首项为[max(n*m)^2]公比为1/4的等比数列前若干项和，具体比例是多少可以求极限算一下。

空间复杂度基本是O(max(n,m)²)的具体数据如下：

Case	total node	input_n	input_m	total rec	rec/node
1	1398101	1024	1024	1048576	75%
2	349525	512	512	262144	75%
3	1398099	1024	512	524288	37.5%
4	349519	500	500	250000	71.53%
5	699051	1024	1	1024	0.1465%

#include <algorithm>
#include <iostream>
#include <cstring>

using namespace std;
const int mxn = 250000;
#define inf 0x3f3f3f3f
inline int son(int p,int x){
    return (p<<2)-2+x;
}
int n,m;
struct Interval:public pair<short ,short >{
#define l first
#define r second
    Interval(short a, short b):pair<short ,short >(a,b){}
    short mid(){
        return ((r-l)>>1)+l;
    }
    short length(){
        return r-l+(short)1;
    }
    Interval left(){
        return Interval(l,mid());
    }
    Interval right(){
        return Interval(mid()+(short)1,r);
    }
    bool in(int L,int R){
        return l <= L and R <= r;
    }
    bool in(const Interval&temp){
        return in(temp.l,temp.r);
    }
    bool intersect(const Interval &k){
        return !( l > k.r || r < k.l );
    }
    void show(){
        cerr<<l<<" "<<r<<"\n";
    }
};
int plant[505][505];
int val[mxn<<2];
void pushUp(int rt){
    for(int i = 0 ; i < 4 ; ++i){
        val[rt] = max(val[rt],val[son(rt,i)]);
    }
}
void buildTree(int rt,Interval x,Interval y){
    if(x.length() <= 0 or y.length() <=0)return;
    if(x.length()==1 and y.length() == 1){
        val[rt] = plant[x.l][y.l];
        return;
    }
    for(int i = 0; i < 4 ; ++i){
        buildTree(son(rt,i) , (i&1)?x.right():x.left(),(i&2)?y.right():y.left());
    }
    pushUp(rt);
}
Interval tarx(0,0),tary(0,0);
int Query(int rt,Interval x,Interval y){
    if(tarx.in(x) and tary.in(y)){
        return val[rt];
    }
    if(!tary.intersect(y) or !tarx.intersect(x))return 0;
    int ret = 0;
    for(int i = 0 ; i < 4 ; ++i){
        ret = max(ret,Query(son(rt,i) , (i&1)?x.right():x.left(),(i&2)?y.right():y.left()));
    }
    return ret;
}
int Query(short begx, short begy,int k){
    tarx = Interval(begx,begx+(short)(k-1));
    tary = Interval(begy,begy+(short)(k-1));
    return Query(1,Interval(1,n),Interval(1,m));
}
int main(){
    ios::sync_with_stdio(false);
    cin>>n>>m;
    int k,t,x,y,a;
    cin>>k>>t;
    memset(plant,0x3f, sizeof(plant));
    while (t--){
        cin>>x>>y>>a;
        plant[x][y] = a;
    }
    int ans = inf;
    buildTree(1,Interval(1,n),Interval(1,m));
    for(int i = 1 ; i <= n-k+1;  ++i){
        for(int j = 1 ; j <= m-k+1; ++j){
            ans = min(ans,Query(i,j,k));
        }
    }
    if(ans!=inf){
        cout<<ans;
    }else cout<<-1;
}