Uva 11297 Census(线段树)

题意：给一个n*n的矩阵，给出q个操作，操作有两种，一种是查询 矩阵内x1 <= x <= x2, y1 <= y <= y2的最大值和最小值，另一种是把(x, y)位置的值改为v

思路：二维线段树的模板题

#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
#include<algorithm>
const int INF = 1 << 30;
const int maxn = 2010;
using namespace std;

int Max[maxn][maxn], Min[maxn][maxn];
int xo, xleaf, x1, y1, x2, y2;
int x, y, v, vmax, vmin, n, m, q;
char op[5];

void query1D(int node, int l, int r) {
if(l >= y1 && r <= y2) {
vmax = max(vmax, Max[xo][node]);
vmin = min(vmin, Min[xo][node]);
} else {
int mid = (l + r) >> 1;
if(y1 <= mid) query1D(node * 2, l, mid);
if(y2 > mid) query1D(node * 2 + 1, mid + 1, r);
}
}

void query2D(int node, int l, int r) {
if(l >= x1 && r <= x2) { xo = node; query1D(1, 1, m); }
else {
int mid = (l + r) >> 1;
if(x1 <= mid) query2D(node * 2, l, mid);
if(x2 > mid) query2D(node * 2 + 1, mid + 1, r);
}
}

void update1D(int node, int l, int r) {
if(l == r) {
if(xleaf) { Max[xo][node] = Min[xo][node] = v; return; }
Max[xo][node] = max(Max[xo * 2][node], Max[xo * 2 + 1][node]);
Min[xo][node] = min(Min[xo * 2][node], Min[xo * 2 + 1][node]);
} else {
int mid = (l + r) >> 1;
if(y <= mid) update1D(node * 2, l, mid);
else update1D(node * 2 + 1, mid + 1, r);
Max[xo][node] = max(Max[xo][node * 2], Max[xo][node * 2 + 1]);
Min[xo][node] = min(Min[xo][node * 2], Min[xo][node * 2 + 1]);
}
}

void update2D(int node, int l, int r) {
if(l == r) {
xo = node; xleaf = 1;
update1D(1, 1, m);
} else {
int mid = (l + r) >> 1;
if(x <= mid) update2D(node * 2, l, mid);
else update2D(node * 2 + 1, mid + 1, r);
xo = node; xleaf = 0;
update1D(1, 1, m);
}
}

int main() {
while(scanf("%d", &n) != EOF) {
m = n;
for(x = 1; x <= n; x++) {
for(y = 1; y <= m; y++) {
scanf("%d", &v);
update2D(1, 1, n);
}
}
scanf("%d", &q);
while(q--) {
scanf("%s", op);
if(op[0] == 'q') {
scanf("%d %d %d %d", &x1, &y1, &x2, &y2);
vmax = -INF; vmin = INF;
query2D(1, 1, n);
printf("%d %d\n", vmax, vmin);
} else {
scanf("%d %d %d", &x, &y, &v);
update2D(1, 1, n);
}
}
}
return 0;
}