1 #include<cstdio> 2 #include<iostream> 3 using namespace std; 4 5 const int maxn = 1005; 6 7 int g[maxn][maxn], ans[maxn][maxn]; 8 int f[4] = {1, 0, -1, 0}; 9 int f1[4] = {0, -1, 0, 1}; 10 int r, c, p; 11 12 inline int dfs(int x, int y){ 13 if(ans[x][y]) return ans[x][y]; 14 int t = 1; 15 for(int i = 0; i <= 3; i++){ 16 int nx = x + f[i]; 17 int ny = y + f1[i]; 18 if(nx >= 1 && ny >= 1 && nx <= r && ny <= c && g[x][y] > g[nx][ny])//是否越界 19 t = max(dfs(nx, ny) + 1, t);//包括自身 20 } 21 ans[x][y] = t;//保存 22 return ans[x][y];//返回 23 } 24 25 int main(){ 26 scanf("%d%d", &r, &c); 27 for(int i = 1; i <= r; i++){ 28 for(int j = 1; j <= c; j++){ 29 scanf("%d", &g[i][j]); 30 } 31 } 32 for(int i = 1; i <= r; i++){ 33 for(int j = 1; j <= c; j++){ 34 ans[i][j] = dfs(i, j); 35 p = max(p, ans[i][j]);//取最大 36 } 37 } 38 printf("%d\n", p); 39 return 0; 40 }
原文:https://www.cnblogs.com/New-ljx/p/11180284.html