3 solutions

  • -3
    @ 2024-11-19 17:07:37
    #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) { //	freopen("number.in", "r", stdin); freopen("number.out", "w", stdout); 	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; }
    

    Information

    ID
    77
    Time
    1000ms
    Memory
    256MiB
    Difficulty
    7
    Tags
    # Submissions
    44
    Accepted
    10
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