题解 | #小美的蛋糕切割#(二维前缀和)

小美的蛋糕切割

https://www.nowcoder.com/practice/15aa2c407c8840e09e2532313fb6809d

#include "bits/stdc++.h"

using namespace std;
#define int long long
#define endl "\n"
#define PII pair<int,int>
#define PIII pair<int,PII>
const int MOD = 1e9 + 7;
const int N = 3e5;

void slu() {
    int n, m;
    cin >> n >> m;
    vector<vector<PIII>> a(n + 1, vector<PIII >(m + 1, {0, {0, 0}}));
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            cin >> a[i][j].first;
            a[i][j].second.first += a[i][j - 1].second.first + a[i][j].first;
            a[i][j].second.second += a[i - 1][j].second.second + a[i][j].first;
            a[i][j].first = a[i][j].second.first + a[i][j].second.second - a[i][j].first + a[i - 1][j - 1].first;
        }
    }
    int dis = INT_MAX;
    int p2 = a[n][m].first;
    for (int i = 0; i <= n; i++) {
        int p1 = a[i][m].first;
        dis = min(dis, abs(p1 - (p2 - p1)));
    }
    for (int i = 0; i <= m; i++) {
        int p1 = a[n][i].first;
        dis = min(dis, abs(p1 - (p2 - p1)));

    }
    cout << dis;
}

signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int T;
//    cin >> T;
    T = 1;
    while (T--)slu();

}

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