Longest Run on a Snowboard Input: standard input Output: standard output Time Limit: 5 seconds Memory Limit: 32 MB Michael likes snowboarding. That's not very surprising, since snowboarding is really great. The bad thing is that in order to gain spee…

Problem C Longest Run on a Snowboard Input: standard input Output: standard output Time Limit: 5 seconds Memory Limit: 32 MB Michael likes snowboarding. That's not very surprising, since snowboarding is really great. The bad thing is that in order to…

Problem C Longest Run on a Snowboard Input: standard input Output: standard output Time Limit: 5 seconds Memory Limit: 32 MB   Michael likes snowboarding. That's not very surprising, since snowboarding is really great. The bad thing is that in order…

题意:和最长滑雪路径一样, #include<iostream> #include<cstdio> #include<cstring> #include <cmath> #include<stack> #include<vector> #include<map> #include<set> #include<queue> #include<algorithm> #define mod=1…

题意: n*m的矩阵,求矩阵中最长下降的序列的长度. 分析: dp[i][j]表示以i,j为起点的最长下降序列,然后记忆化搜索. #include <map> #include <set> #include <list> #include <cmath> #include <queue> #include <stack> #include <cstdio> #include <vector> #include…

称号:给你一个二维矩阵,找到一个点.每一个可以移动到的位置相邻的上下,求最长单调路径. 分析:贪婪,dp.搜索. 这个问题是一个小样本,我们该怎么办. 这里使用贪心算法: 首先.将全部点依照权值排序(每一个点一定被值更大的点更新): 然后,按顺序更新排序后.每一个点更新周围的点: 最后,找到最大值输出就可以. 说明:╮(╯▽╰)╭居然拍了1000+,还以为这样的方法比較快呢(数据分布啊╮(╯▽╰)╭). #include <algorithm> #include <iostream>…

思路:d[x][y]表示以(x, y)作为起点能得到的最长递减序列,转移方程d[x][y] = max(d[px][py] + 1),此处(px, py)是它的相邻位置并且该位置的值小于(x, y)处的值.可以选择把所有坐标根据值的大小升序排序,因为值较大的坐标取决于值更小的相邻坐标.   AC代码: #include<cstdio> #include<algorithm> #include<cstring> #include<utility> #inclu…

记忆化搜索,完事... Code /** * UVa * Problem#10285 * Accepted * Time:0ms */ #include<iostream> #include<fstream> #include<sstream> #include<algorithm> #include<cstdio> #include<cstring> #include<cstdlib> #include<cctyp…

题目大意:滑雪.给一个二维数组,找出最长的连续下降序列的长度. 题目分析:定义dp(i,j)表示以a[i][j]结尾的最长连续下降序列的长度,则dp(i,j)=max(dp(i-1,j),dp(i+1,j),dp(i,j-1),dp(i,j+1)),重复这个DP过程滚动更新dp数组100次即可(我专门做了一下测试,更新次数在42次及以上时都能AC). 代码如下: # include<iostream> # include<cstdio> # include<string>…

题意 其实就是一条二维的LIS,但是还是做的一愣一愣的,多努力. 考虑$dp[i][j]$为从(i,j)出发的二维LIS的最大值,那么$dp[i][j]=max\{dp[i−di[k]][j−dj[k]]\}+1$,取dp值时要求严格递减.否则值为1. 分析 #include <bits/stdc++.h> #define MP make_pair #define PB push_back #define fi first #define se second #define ZERO(x) m…