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最新推荐文章于 2016-12-05 16:37:39 发布

原创最新推荐文章于 2016-12-05 16:37:39 发布 · 425 阅读

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树状数组

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本文介绍了一种使用三维树状数组解决立方体元素求和问题的方法。通过维护一个三维树状数组，可以高效地更新立方体中任意位置的数值，并快速获取从坐标(x1,y1,z1)到(x2,y2,z2)的元素总和。

维护一个三维树状数组即可。
唯一的难点在于如何分解立方体从而得到从(x1, y1, z1)到(x2, y2, z2)的元素和。

#include <cstdio>

const int maxn = 128 + 5;

long long bit[maxn][maxn][maxn];

int n;

int lowbit(int x) {
    return x & -x;
}

void add(int x, int y, int z, int delta) {
    for (int i = x; i <= n + 1; i += lowbit(i)) {
        for (int j = y; j <= n + 1; j += lowbit(j)) {
            for (int k = z; k <= n + 1; k += lowbit(k)) {
                bit[i][j][k] += delta;
            }
        }
    }
}

int sum(int x, int y, int z) {
    int ret = 0;
    for (int i = x; i > 0; i -= lowbit(i)) {
        for (int j = y; j > 0; j -= lowbit(j)) {
            for (int k = z; k > 0; k -= lowbit(k)) {
                ret += bit[i][j][k];
            }
        }
    }
    return ret;
}

int main(int argc, char const *argv[]) {
    int m;
    scanf("%d", &n);
    while (scanf("%d", &m) == 1 && m != 3) {
        if (m == 1) {
            int x, y, z, k;
            scanf("%d%d%d%d", &x, &y, &z, &k);
            x++, y++, z++;
            add(x, y, z, k);
        } else {
            int x1, y1, z1, x2, y2, z2;
            scanf("%d%d%d%d%d%d", &x1, &y1, &z1, &x2, &y2, &z2);
            x1++, y1++, z1++;
            x2++, y2++, z2++;
            int ans = sum(x2, y2, z2) - sum(x2, y1-1, z2) - sum(x1-1, y2, z2)
            + sum(x1-1, y1-1, z2) - sum(x2, y2, z1-1) + sum(x2, y1-1, z1-1)
            + sum(x1-1, y2, z1-1) - sum(x1-1, y1-1, z1-1);
            printf("%d\n", ans);
        }
    }
    return 0;
}