版权声明:如需转载,记得标识出处 https://blog.csdn.net/godleaf/article/details/87989300
题目链接:https://vjudge.net/problem/POJ-2019
我写的这个二维线段树没什么技巧,就是直接开一个 node[cur][cure],cur代表的是x轴的一维线段树,cure代表的是y轴的一维线段树。
初始化:
先从x轴开始做线段树的操作,当x轴到达叶节点时,再对y轴做线段树的操作,x轴返回到父节点时,一样要对y轴从1到N重头做一次线段树的操作,这是初始化的过程,总的时间是 log(n)*log(n)。也就是说,对每一个x轴节点都要从 1到N做一次y轴的线段树操作。
初始化的这部分我一开始贪简便,把x轴和y轴的线段树操作都写在一个自定义函数里面,导致程序在递归的过程中走了很多重复的步骤。所以对于x轴的线段树操作和y轴的线段树操作要尽量分开写。
查找:
查找的大致流程和初始化是差不多的,先把x轴的取值范围确定下来,然后再去确定y轴的取值范围。
#include <iostream>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
using namespace std;
#define lson (cur<<1)
#define rson (cur<<1|1)
typedef long long ll;
typedef pair<int, int> pii;
const int INF = 0x3f3f3f3f;
const long long LINF = 1e18;
const int Maxn = 250+10;
const int Mod = 1e9+7;
struct Node {
int maxn, minx;
} node[Maxn<<2][Maxn<<2];
int a[Maxn][Maxn], N;
void pushup(int cur, int cure) {
node[cur][cure].maxn = max(node[cur][cure<<1].maxn, node[cur][cure<<1|1].maxn);
node[cur][cure].minx = min(node[cur][cure<<1].minx, node[cur][cure<<1|1].minx);
}
void updata_(int cur, int cure, int l, int r, int L, int R) {
if(l == r && L == R) {
node[cur][cure].maxn = node[cur][cure].minx = a[l][L];
return;
}
if(L == R) {
node[cur][cure].maxn = max(node[cur<<1][cure].maxn, node[cur<<1|1][cure].maxn);
node[cur][cure].minx = min(node[cur<<1][cure].minx, node[cur<<1|1][cure].minx);
return;
}
int Mid = (L+R)>>1;
updata_(cur, cure<<1, l, r, L, Mid);
updata_(cur, cure<<1|1, l, r, Mid+1, R);
pushup(cur, cure);
}
void updata(int cur, int cure, int l, int r, int L, int R) {
int mid = (l+r)>>1;
if(l != r) {
updata(cur<<1, cure, l, mid, L, R);
updata(cur<<1|1, cure, mid+1, r, L, R);
}
updata_(cur, cure, l, r, L, R);
}
int query(int cur, int cure, int l, int r, int L, int R, int x, int y, int B) {
int xx = x+B-1, yy = y+B-1;
if(x <= l && r <= xx && y <= L && R <= yy) return node[cur][cure].maxn;
int mid = (l+r)>>1, Mid = (L+R)>>1, maxn = -INF;
if(x <= l && r <= xx) {
if(y <= Mid) maxn = max(maxn, query(cur, cure<<1, l, r, L, Mid, x, y, B));
if(Mid+1 <= yy) maxn = max(maxn, query(cur, cure<<1|1, l, r, Mid+1, R, x, y, B));
} else {
if(x <= mid) maxn = max(maxn, query(cur<<1, cure, l, mid, L, R, x, y, B));
if(mid+1 <= xx) maxn = max(maxn, query(cur<<1|1, cure, mid+1, r, L, R, x, y, B));
}
return maxn;
}
int Query(int cur, int cure, int l, int r, int L, int R, int x, int y, int B) {
int xx = x+B-1, yy = y+B-1;
if(x <= l && r <= xx && y <= L && R <= yy) return node[cur][cure].minx;
int mid = (l+r)>>1, Mid = (L+R)>>1, minx = INF;
if(x <= l && r <= xx) {
if(y <= Mid) minx = min(minx, Query(cur, cure<<1, l, r, L, Mid, x, y, B));
if(Mid+1 <= yy) minx = min(minx, Query(cur, cure<<1|1, l, r, Mid+1, R, x, y, B));
} else {
if(x <= mid) minx = min(minx, Query(cur<<1, cure, l, mid, L, R, x, y, B));
if(mid+1 <= xx) minx = min(minx, Query(cur<<1|1, cure, mid+1, r, L, R, x, y, B));
}
return minx;
}
int main(void)
{
int B, K;
scanf("%d%d%d", &N, &B, &K);
for(int i = 0; i < N; ++i)
for(int j = 0; j < N; ++j) scanf("%d", &a[i][j]);
updata(1, 1, 0, N-1, 0, N-1);
int x, y;
while (K--) {
scanf("%d%d", &x, &y);
x--; y--;
int maxn, minx;
maxn = query(1, 1, 0, N-1, 0, N-1, x, y, B);
minx = Query(1, 1, 0, N-1, 0, N-1, x, y, B);
printf("%d\n", maxn-minx);
}
return 0;
}