HDU 1892 See you~(二维树状数组的单点更新，区间求值)

题目链接：点击打开链接

二维树状数组单点更新，区间求值的入门题目：

一维树状数组的思想很容易应用到二维树状数组中，sum[i][j]表示一块矩阵的和，其中i - lowBit(i) < x <= i，j - lowBit(j) < y <= j。

那么求以(x1，y1) （左下角） 和 (x2，y2)（右上角） 为对角线的矩阵的和为getSum(x2，y2) - getSum(x1 - 1，y2) - getSum(x2，y1 - 1) + getSum(x1 - 1，y1 - 1)。

// HDU 1892 See you~.cpp 运行：1248ms#include "stdafx.h"#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>using namespace std;int sum[1005][1005];int lowBit(int x) {return x & (-x);}void update(int x, int y, int n) {for (int i = x; i < 1005; i += lowBit(i)) {for (int j = y; j < 1005; j += lowBit(j)) {sum[i][j] += n;}}}int getSum(int x, int y) {int re = 0;for (int i = x; i > 0; i -= lowBit(i)) {for (int j = y; j > 0; j -= lowBit(j)) {re += sum[i][j];}}return re;}int getValue(int x, int y) {//单点更新，区间求值的树状数组也可以单点求值return getSum(x, y) - getSum(x - 1, y) - getSum(x, y - 1) + getSum(x - 1,y - 1);}void init() {memset(sum, 0, sizeof(sum));for (int i = 1; i < 1005; i++) {for (int j = 1; j < 1005; j++) {update(i, j, 1);}}}int main(){int k,m;int x1, y1, x2, y2, n1;//题目给的坐标为从0计数，树状数组需要从下标1开始计数，因此操作时需要转换一下int x11, y11, x22, y22, value;scanf("%d", &k);for(int i = 1;i <= k;i++){scanf("%d", &m);printf("Case %d:\n", i);init();for (int j = 0; j < m; j++) {getchar();char ch = getchar();switch (ch) {case 'S':scanf("%d%d%d%d", &x1, &y1, &x2, &y2);x11 = max(x1, x2);x22 = min(x1, x2);y11 = max(y1, y2);y22 = min(y1, y2);printf("%d\n", getSum(x11 + 1, y11 + 1) - getSum(x22, y11 + 1) - getSum(x11 + 1, y22) + getSum(x22, y22));break;case 'A':scanf("%d%d%d", &x1, &y1, &n1);update(x1 + 1, y1 + 1, n1);break;case 'D':scanf("%d%d%d", &x1, &y1, &n1);value = getValue(x1 + 1,y1 + 1);n1 = min(value, n1);update(x1 + 1, y1 + 1, -n1);break;case 'M':scanf("%d%d%d%d%d", &x1, &y1, &x2, &y2, &n1);value = getValue(x1 + 1, y1 + 1);n1 = min(value, n1);update(x1 + 1, y1 + 1, -n1);update(x2 + 1, y2 + 1, n1);break;}}}    return 0;}