1722E - Counting Rectangles - CodeForces Solution

brute force data structures dp implementation

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Python Code: 

def solve(rectangles, queries):
    acc = [[0] * 1001 for _ in range(1001)]
    for hi, wi in rectangles:
        acc[hi][wi] += hi * wi

    for r in range(1001):
        for c in range(1001):
            acc[r][c] += acc[r-1][c] + acc[r][c-1] - acc[r-1][c-1]

    ans = []
    for hs, ws, hb, wb in queries:
        cur = acc[hb-1][wb-1] - acc[hb-1][ws] - acc[hs][wb-1] + acc[hs][ws]
        ans.append(cur)

    return ans


def main():
    t = int(input())
    output = []
    for _ in range(t):
        n, q = read_ints()
        rectangles = [read_ints() for _ in range(n)]
        queries = [read_ints() for _ in range(q)]

        ans = solve(rectangles, queries)
        output += ans

    print_int_lines(output)


def read_ints():
    return [int(c) for c in input().split()]


def print_int_lines(lst):
    print('\n'.join(map(str, lst)))


if __name__ == "__main__":
    main()

C++ Code: 

#include <bits/stdc++.h>
# define int long long
using namespace std;

const int cnst = 1e3+5;
// const int mod = 1e9+7;
bool mutipletestcase = true;

void solve() {
    int n, q; 
    cin >> n >> q;

    int presum2d[cnst][cnst];
    memset(presum2d, 0, sizeof(presum2d));

    for(int i = 1; i<=n; i++) {
        int a, b; cin >> a >> b;
        presum2d[a][b] += (a*b);
    }


    for(int i = 1; i<=cnst-5; i++) {
        for(int j = 1; j<=cnst-5; j++) {
            presum2d[i][j] += presum2d[i-1][j]+presum2d[i][j-1]-presum2d[i-1][j-1];
        }
    }

    while(q--) {
        int rst, cst, rnd, cnd; cin >> rst >> cst >> rnd >> cnd;
        cout << presum2d[rnd-1][cnd-1]-presum2d[rst][cnd-1]-presum2d[rnd-1][cst]+presum2d[rst][cst] << endl;
    }
}

signed main() {
    ios_base::sync_with_stdio(false);
    int t = 1; 
    if(mutipletestcase) cin >> t;
    while(t--) solve();
}


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