题解Luogu P2940 【[USACO09FEB]音乐妖精The Leprechaun】

\(O(n^2)\)算法

把矩阵分成好多个环，用O(n)时间求环上最大连续子段和(对最大连续子段和以及sum减去最小连续子段和取最大值)

注意要特判全部是负数的情况(最大连续子段和可能会一个都不取)

#include <bits/stdc++.h>

using namespace std;

const int N = 200;
int ma[N << 1][N << 1];

int main()
{
    int n, m, ans = -1E9;
    scanf("%d", &n);
    m = n << 1;
    for (int i = 0; i < n; ++i)
        for (int j = 0; j < n; ++j)
        {
            scanf("%d", &ma[i][j]);
            ans = max(ans, ma[i][j]);
            ma[i + n][j] = ma[i][j + n] = ma[i + n][j + n] = ma[i][j];
        }
    if (ans <= 0)
    {
        printf("%d", ans);
        return 0;
    }
    for (int i = 0; i < n; ++i)
    {
        int a = 0, b = 0, c = 0, d = 0, e = 0, f = 0, g = 0, h = 0, tmp1 = 0,
            tmp2 = 0, tmp3 = 0, tmp4 = 0, sum1 = 0, sum2 = 0, sum3 = 0,
            sum4 = 0;
        for (int j = 0; j < n; ++j)
        {
            sum1 += ma[i][j];
            a += ma[i][j];
            if (a < 0) a = 0;
            sum2 += ma[j][i];
            b += ma[j][i];
            if (b < 0) b = 0;
            c += ma[i][j];
            if (c > 0) c = 0;
            tmp1 = min(tmp1, c);
            d += ma[j][i];
            if (d > 0) d = 0;
            tmp2 = min(tmp2, d);
            sum3 += ma[i + j][j];
            e += ma[i + j][j];
            if (e < 0) e = 0;
            sum4 += ma[j][i + j];
            f += ma[j][i + j];
            if (f < 0) f = 0;
            g += ma[i + j][j];
            if (g > 0) g = 0;
            tmp3 = min(tmp3, g);
            h += ma[j][i + j];
            if (h > 0) h = 0;
            tmp4 = min(tmp4, h);
            ans = max(ans, max(max(a, b), max(e, f)));
        }
        ans = max(ans, max(max(sum1 - tmp1, sum2 - tmp2),
                           max(sum3 - tmp3, sum4 - tmp4)));
    }
    for (int i = n - 1; i < m - 1; ++i)
    {
        int a = 0, b = 0, tmp = 0, sum = 0;
        for (int j = 0; j < n; ++j)
        {
            sum += ma[i - j][j];
            a += ma[i - j][j];
            if (a < 0) a = 0;
            b += ma[i - j][j];
            if (b > 0) b = 0;
            tmp = min(tmp, b);
            ans = max(ans, a);
        }
        ans = max(ans, sum - tmp);
    }
    printf("%d", ans);
    return 0;
}