【动态规划】 51nod 1084 矩阵来回取数

通道

思路:dp[i][j][k]:k步时，A到i行，B到j行的最大值

代码：

#include <cstdio>
#include <vector>
using namespace std;

int maxPath(vector<vector<int> > &num) {
    const int m = num.size();
    const int n = num[0].size();
    const int l = m + n - 1;
    vector<vector<int> > g(m + 1, vector<int>(m + 1));
    vector<vector<int> > h(m + 1, vector<int>(m + 1));
    for (int k = 1; k <= l; k++) {
        int st = k <= n ? 1 : k - n + 1;
        int ed = k <= m ? k : m;
        for (int i = st; i <= ed; i++) 
            for (int j = i; j <= ed; j++) {
                h[i][j] = num[i - 1][k - i] +max(max(g[i - 1][j - 1], g[i - 1][j]), max(g[i][j - 1], g[i][j]));
                if (i != j) 
                    h[i][j] += num[j - 1][k - j];
            }
        swap(g, h);
    }
    return g[m][m];
}

int main() {
    int m, n;
    scanf("%d%d", &m, &n);
    vector<vector<int> > num(n, vector<int>(m));
    for (int i = 0; i < n; i++) 
        for (int j = 0; j < m; j++) 
            scanf("%d", &num[i][j]);
    printf("%d\n", maxPath(num));
    return 0;
}

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转载于:https://www.cnblogs.com/Rojo/p/4852666.html