POJ 2019 Cornfields 【二维线段树】

题目链接：POJ 2019 Cornfields

Cornfields
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 6269 Accepted: 3089
Description

FJ has decided to grow his own corn hybrid in order to help the cows make the best possible milk. To that end, he’s looking to build the cornfield on the flattest piece of land he can find.

FJ has, at great expense, surveyed his square farm of N x N hectares (1 <= N <= 250). Each hectare has an integer elevation (0 <= elevation <= 250) associated with it.

FJ will present your program with the elevations and a set of K (1 <= K <= 100,000) queries of the form “in this B x B submatrix, what is the maximum and minimum elevation?”. The integer B (1 <= B <= N) is the size of one edge of the square cornfield and is a constant for every inquiry. Help FJ find the best place to put his cornfield.
Input

• Line 1: Three space-separated integers: N, B, and K.

• Lines 2..N+1: Each line contains N space-separated integers. Line 2 represents row 1; line 3 represents row 2, etc. The first integer on each line represents column 1; the second integer represents column 2; etc.

• Lines N+2..N+K+1: Each line contains two space-separated integers representing a query. The first integer is the top row of the query; the second integer is the left column of the query. The integers are in the range 1..N-B+1.
  Output

• Lines 1..K: A single integer per line representing the difference between the max and the min in each query.
  Sample Input

5 3 1
5 1 2 6 3
1 3 5 2 7
7 2 4 6 1
9 9 8 6 5
0 6 9 3 9
1 2
Sample Output

5

AC代码：

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <algorithm>#include <vector>#include <queue>#include <map>#include <stack>#define PI acos(-1.0)#define CLR(a, b) memset(a, (b), sizeof(a))#define fi first#define se second#define ll o<<1#define rr o<<1|1using namespace std;typedef long long LL;typedef pair<int, int> pii;const int MAXN = 250 + 10;const int pN = 1e6;// <= 10^7const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;void add(LL &x, LL y) { x += y; x %= MOD; }struct Nodey {    int ly, ry, Max, Min;};int nx, ny;int posx[MAXN], posy[MAXN];struct Nodex {    int lx, rx;    Nodey treey[MAXN<<2];    void Build_y(int o, int l, int r) {        treey[o].ly = l; treey[o].ry = r;        treey[o].Max = 0; treey[o].Min = 300;        if(l == r) {            posy[l] = o;            return ;        }        int mid = (l + r) >> 1;        Build_y(ll, l, mid);        Build_y(rr, mid+1, r);    }    int Query_y(int o, int y1, int y2, int op) {        if(treey[o].ly == y1 && treey[o].ry == y2) {            return op ? treey[o].Max : treey[o].Min;        }        int mid = (treey[o].ly + treey[o].ry) >> 1;        if(y2 <= mid) return Query_y(ll, y1, y2, op);        else if(y1 > mid) return Query_y(rr, y1, y2, op);        else return op ? max(Query_y(ll, y1, mid, op), Query_y(rr, mid+1, y2, op))                       : min(Query_y(ll, y1, mid, op), Query_y(rr, mid+1, y2, op));    }};Nodex treex[MAXN<<2];void Build_x(int o, int l, int r) {    treex[o].lx = l; treex[o].rx = r;    treex[o].Build_y(1, 1, ny);    if(l == r) {        posx[l] = o;        return ;    }    int mid = (l + r) >> 1;    Build_x(ll, l, mid);    Build_x(rr, mid+1, r);}int Query_x(int o, int x1, int x2, int y1, int y2, int op) {    if(treex[o].lx == x1 && treex[o].rx == x2) {        return treex[o].Query_y(1, y1, y2, op);    }    int mid = (treex[o].lx + treex[o].rx) >> 1;    if(x2 <= mid) return Query_x(ll, x1, x2, y1, y2, op);    else if(x1 > mid) return Query_x(rr, x1, x2, y1, y2, op);    else return op ? max(Query_x(ll, x1, mid, y1, y2, op), Query_x(rr, mid+1, x2, y1, y2, op))                   : min(Query_x(ll, x1, mid, y1, y2, op), Query_x(rr, mid+1, x2, y1, y2, op));}void PushUp(int x, int y, int v) {    treex[x].treey[y].Max = max(treex[x].treey[y].Max, v);    treex[x].treey[y].Min = min(treex[x].treey[y].Min, v);}void Update(int x, int y, int v) {    for(int i = posx[x]; i ; i >>= 1) {        for(int j = posy[y]; j ; j >>= 1) {            PushUp(i, j, v);        }    }}int main(){    int b, m;    scanf("%d%d%d", &nx, &b, &m); ny = nx;    Build_x(1, 1, nx);    for(int i = 1; i <= nx; i++) {        for(int j = 1; j <= ny; j++) {            int v; scanf("%d", &v);            Update(i, j, v);        }    }    while(m--) {        int x, y; scanf("%d%d", &x, &y);        printf("%d\n", Query_x(1, x, x + b - 1, y, y + b - 1, 1) - Query_x(1, x, x + b - 1, y, y + b - 1, 0));    }    return 0;}