AcWing 1232. 三体攻击（三维差分）

差分和前缀和互为逆运算，这个题用三维差分，三维只是比较麻烦，和容斥原理比较类似。

Problem
好像是A组第六题，蓝桥杯省赛第六题居然已经这么难了？

因为保证矩阵行列相乘小于1e6但不知道行和列分别有多大，所以需要用到数组下标映射，也就是代码中的

public static int get(int i, int j, int k){...}

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;

class Main {
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static PrintWriter pw = new PrintWriter(System.out);
    static int N = 2000010, A, B, C, m;
    static long s[] = new long[N], b[] = new long[N], bp[] = new long[N];
    static int op[][] = new int[N >> 1][7];
    static int d[][] = new int[][]{
            {0, 0, 0, 1},
            {0, 0, 1, -1},
            {0, 1, 0, -1},
            {0, 1, 1, 1},
            {1, 0, 0, -1},
            {1, 0, 1, 1},
            {1, 1, 0, 1},
            {1, 1, 1, -1},
    };

    public static void main(String[] args) throws Exception {

        String ss[] = br.readLine().split(" ");
        A = Integer.parseInt(ss[0]);
        B = Integer.parseInt(ss[1]);
        C = Integer.parseInt(ss[2]);
        m = Integer.parseInt(ss[3]);

        ss = br.readLine().split(" ");

        int uu = 0;
        for (int i = 1; i <= A; i++) {
            for (int j = 1; j <= B; j++) {
                for (int k = 1; k <= C; k++) {
                    s[get(i, j, k)] = Long.parseLong(ss[uu++]);
                }
            }
        }

        for (int i = 1; i <= A; i++) {
            for (int j = 1; j <= B; j++) {
                for (int k = 1; k <= C; k++) {
                    for (int u = 0; u < 8; u++) {
                        int x = i - d[u][0];
                        int y = j - d[u][1];
                        int z = k - d[u][2];
                        int t = d[u][3];
                        bp[get(i, j, k)] += s[get(x, y, z)] * t;
                    }
                }
            }
        }

        for (int i = 1; i <= m; i++) {
            ss = br.readLine().split(" ");
            for (int j = 0; j < 7; j++)
                op[i][j] = Integer.parseInt(ss[j]);
        }

        int l = 1, r = m;
        while (l < r) {
            int mid = l + r >> 1;
            if (check(mid)) r = mid;
            else l = mid + 1;
        }

        pw.print(r);
        pw.flush();
        pw.close();
        br.close();

    }

    public static int get(int i, int j, int k) {
        return (i * B + j) * C + k;
    }

    public static boolean check(int mid) {
        b = Arrays.copyOf(bp, bp.length);
        for (int i = 1; i <= mid; i++) {
            int x1 = op[i][0];
            int x2 = op[i][1];
            int y1 = op[i][2];
            int y2 = op[i][3];
            int z1 = op[i][4];
            int z2 = op[i][5];
            int t = op[i][6];
            b[get(x1    , y1    , z1    )] -= t;
            b[get(x1    , y1    , z2 + 1)] += t;
            b[get(x1    , y2 + 1, z1    )] += t;
            b[get(x1    , y2 + 1, z2 + 1)] -= t;
            b[get(x2 + 1, y1    , z1    )] += t;
            b[get(x2 + 1, y1    , z2 + 1)] -= t;
            b[get(x2 + 1, y2 + 1, z1    )] -= t;
            b[get(x2 + 1, y2 + 1, z2 + 1)] += t;
        }

        Arrays.fill(s, 0);
        for (int i = 1; i <= A; i++) {
            for (int j = 1; j <= B; j++) {
                for (int k = 1; k <= C; k++) {
                    s[get(i, j, k)] = b[get(i, j, k)];
                    for (int u = 1; u < 8; u++) {
                        int x = i - d[u][0];
                        int y = j - d[u][1];
                        int z = k - d[u][2];
                        int t = d[u][3];
                        s[get(i, j, k)] -= s[get(x, y, z)] * t;
                    }
                    if (s[get(i, j, k)] < 0) return true;
                }
            }
        }
        return false;
    }

}

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来源： https://blog.csdn.net/qq_43515011/article/details/104516245