随笔分类 -  动态规划——最大子矩阵

摘要:好的题解使人一下就懂啊…… s1[i][j]表示(i,j)最多向左(或右)延伸多少个格子,使这些格子中的数都是0(不包括(i,j)) s2[i][j]表示(i,j)最多向上延伸多少个格子,使这些格子中的数都是0(不包括(i,j)) f[i][j]表以(i,j)为右下(左下)角的最大对角线长度 来自 阅读全文
posted @ 2017-12-18 11:40 poorpool 阅读(139) 评论(0) 推荐(0)
摘要:悬线法 cpp include include include using namespace std; int n, m, h[2005][2005], l[2005][2005], r[2005][2005], uu, iii=0, jjj=0; bool a[2005][2005]; void 阅读全文
posted @ 2017-12-13 19:23 poorpool 阅读(134) 评论(0) 推荐(0)
摘要:悬线法 cpp include include include using namespace std; int n, m, ans=0, h[1005][1005], l[1005][1005], r[1005][1005]; bool a[1005][1005]; char s[15]; int 阅读全文
posted @ 2017-12-13 18:41 poorpool 阅读(124) 评论(0) 推荐(0)
摘要:建议看看 "王知昆dalao的论文" ,讲得很好 cpp include include include include using namespace std; int l, w, n, ans=0, sh, xi, ma; struct Node{ int x, y; }nd[5105]; bo 阅读全文
posted @ 2017-12-13 10:48 poorpool 阅读(136) 评论(0) 推荐(0)