1 条题解

  • 5
    @ 2023-8-26 9:29:01

    累死我了 看在我这么辛苦,点个赞吧😄

    #include <stdio.h>
    typedef long long LL;
    const LL min_ll = -1e18;
    int n, m; LL w[1005][1005], f[1005][1005][2];
    inline LL mx(LL p, LL q, LL r) {return p > q ? (p > r ? p : r) : (q > r ? q : r);}
    inline LL dfs(int x, int y, int from) {
        if (x < 1 || x > n || y < 1 || y > m) return min_ll;
        if (f[x][y][from] != min_ll) return f[x][y][from];
        if (from == 0) f[x][y][from] = mx(dfs(x + 1, y, 0), dfs(x, y - 1, 0), dfs(x, y - 1, 1)) + w[x][y];
        else f[x][y][from] = mx(dfs(x - 1, y, 1), dfs(x, y - 1, 0), dfs(x, y - 1, 1)) + w[x][y];
        return f[x][y][from];
    }
    int main(void) {
    	scanf("%d %d", &n, &m);
    	for (int i = 1; i <= n; ++i)
    		for (int j = 1; j <= m; ++j) {
    			scanf("%lld", &w[i][j]);
    			f[i][j][0] = f[i][j][1] = min_ll;
    		}
        f[1][1][0] = f[1][1][1] = w[1][1];
    	printf("%lld\n", dfs(n, m, 1));
    	return 0;
    }
    
    • 1

    信息

    ID
    1353
    时间
    1000ms
    内存
    256MiB
    难度
    5
    标签
    递交数
    120
    已通过
    49
    上传者