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样例
给定一个矩阵
[
[1 ,2 ,3 ,4 ,5],
[16,17,24,23,6],
[15,18,25,22,7],
[14,19,20,21,8],
[13,12,11,10,9]
]
返回 25
思路：记忆化搜索 + dp
设Lics(num)表示以num开头的最长上升子连续序列的长度， 则Lics(A[x][y]) = max(Lics(A[x-1][y]), Lics(A[x][y-1]), Lics(x+1,y), Lics(x, y+1))+1;
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<vector> #include<deque> #include<map> using namespace std; class Solution { public: /** * @param A an integer matrix * @return an integer */ int n, m, maxL; int lics[1000][1000]; int vis[1000][1000]; int dir[4][2] = {{0, 1}, {1, 0}, {0,-1}, {-1,0}}; int dfs(vector<vector<int>>& A, int x, int y){ int maxLics = 0; vis[x][y] = 1; for(int i=0; i<4; ++i){ int xx = x+dir[i][0]; int yy = y+dir[i][1]; if(xx<0 || yy<0 || xx>=n || yy>=m) continue; if(A[x][y] >= A[xx][yy]) continue; if(!vis[xx][yy]) maxLics = max(maxLics, dfs(A, xx, yy)); else maxLics = max(maxLics, lics[xx][yy]); } lics[x][y] = maxLics+1; if(maxL < lics[x][y]) maxL = lics[x][y]; return lics[x][y]; } int longestIncreasingContinuousSubsequenceII(vector<vector<int>>& A) { n = A.size(); if(n == 0) return 0; m = A[0].size(); memset(lics, 0, sizeof(lics)); memset(vis, 0, sizeof(vis)); maxL = 0; for(int i=0; i<n; ++i) for(int j=0; j<m; ++j) if(!vis[i][j]) dfs(A, i, j); return maxL; } }; /* 1 2 3 4 5 16 17 24 23 6 15 18 25 22 7 14 19 20 21 8 13 12 11 10 9 */
lintcode 最长上升连续子序列 II(二维最长上升连续序列)
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原文地址：http://www.cnblogs.com/hujunzheng/p/4986801.html
