【蓝桥备赛】肖恩的投球游戏加强版——基础二维差分

题目链接

肖恩的投球游戏加强版

个人思路

q 次对一个 n * m 的矩阵操作，我们肯定不能对这个矩阵每次都进行修改，不然复杂度就是 O (q n m) 也就是 1e11， 明显会超时的。

那么，我们就需要拿一个数组记录下这 q 次操作都干了什么？

	while(q--)
    {
        int x1, y1, x2, y2;
        ll c;
        cin >> x1 >> y1 >> x2 >> y2 >> c;
        brr[x1][y1] += c;
        brr[x1][y2 + 1] -= c;
        brr[x2 + 1][y1] -= c;
        brr[x2 + 1][y2 + 1] += c;
    }
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j<= m; ++j)
            brr[i][j] += brr[i - 1][j] + brr[i][j - 1] - brr[i - 1][j - 1];

这里brr数组初始是一个空的数组。在这里我们将这 q 次进行标记，并通过求前缀和获得对于每一个单元格的操作结果，到底是 +10 了还是 -4 了，都清晰的显示在 brr 数组上。
之后，我们就可以将 arr数组 + brr数组 得到修改后的 arr 数组。

	for(int i = 1; i <= n; ++i)
        for(int j = 1; j<= m; ++j)
            arr[i][j] += brr[i][j];

参考代码

C/C++

#include
using namespace std;
typedef long long ll;
const int N = 1e3 + 10;
int n, m, q;
ll arr[N][N], brr[N][N];
int main()
{
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> n >> m >> q;
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j<= m; ++j)
            cin >> arr[i][j];
    while(q--)
    {
        int x1, y1, x2, y2;
        ll c;
        cin >> x1 >> y1 >> x2 >> y2 >> c;
        brr[x1][y1] += c;
        brr[x1][y2 + 1] -= c;
        brr[x2 + 1][y1] -= c;
        brr[x2 + 1][y2 + 1] += c;
    }
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j<= m; ++j)
            brr[i][j] += brr[i - 1][j] + brr[i][j - 1] - brr[i - 1][j - 1];
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j<= m; ++j)
            arr[i][j] += brr[i][j];
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j<= m; ++j)
            cout << arr[i][j] << " \n"[j == m];
}

Java

import java.io.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner();
        int n = sc.nextInt();
        int m = sc.nextInt();
        int q = sc.nextInt();
        long[][] arr = new long[n + 5][m + 5];
        long[][] brr = new long[n + 5][m + 5];
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= m; ++j)
                arr[i][j] = sc.nextLong();
        while (q-- > 0) {
            int x1 = sc.nextInt();
            int y1 = sc.nextInt();
            int x2 = sc.nextInt();
            int y2 = sc.nextInt();
            long c = sc.nextLong();
            brr[x1][y1] += c;
            brr[x1][y2 + 1] -= c;
            brr[x2 + 1][y1] -= c;
            brr[x2 + 1][y2 + 1] += c;
        }
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= m; ++j)
                brr[i][j] += brr[i - 1][j] + brr[i][j - 1] - brr[i - 1][j - 1];
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= m; ++j)
                arr[i][j] += brr[i][j];
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= m; ++j) {
                System.out.print(arr[i][j]);
                System.out.print(j == m ? "\n" : " ");
            }

    }
}

class Scanner {
    static StreamTokenizer st = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));

    public long nextLong() {
        try {
            st.nextToken();
        } catch (IOException e) {
            throw new RuntimeException(e);
        }
        return (long) st.nval;
    }

    public int nextInt() {
        try {
            st.nextToken();
        } catch (IOException e) {
            throw new RuntimeException(e);
        }
        return (int) st.nval;
    }
}