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给一个S*S的矩阵，有两种操作，给(x,y)位置增加一个值，和求一个内部矩形的和。
二维树状数组，先对每行来一个一维的树状数组，
再对行来一个树状数组
#include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std;typedef long long LL;const int MAXN = 2000;int c[MAXN][MAXN];int n, op;int Lowbit(int x){ return x&(-x);}void Update(int x, int y, int v){ for (int i = x;i <= n;i += Lowbit(i)) { for (int j = y;j <= n;j += Lowbit(j)) { c[i][j] += v; } }}int Sum(int x, int y){ int res = 0; for (int i = x;i > 0;i -= Lowbit(i)) { for (int j = y;j > 0;j -= Lowbit(j)) { res += c[i][j]; } } return res;}int main(){ scanf("%d%d", &op, &n); while (~scanf("%d", &op)) { if (op == 3) break; if (op == 1) { int x, y, v; scanf("%d%d%d", &x, &y, &v); x++, y++; Update(x, y, v); } if (op == 2) { int x1, y1, x2, y2; scanf("%d%d%d%d", &x1, &y1, &x2, &y2); x1++, y1++, x2++, y2++; printf("%d\n", Sum(x2, y2) - Sum(x1-1, y2) - Sum(x2, y1-1) + Sum(x1-1, y1-1)); } } return 0;}
转载于:https://www.cnblogs.com/YDDDD/p/10700465.html