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…

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…

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…

题意:在一个R*C(R, C<=100)的整数矩阵上找一条高度严格递减的最长路.起点任意,但每次只能沿着上下左右4个方向之一走一格,并且不能走出矩阵外.矩阵中的数均为0~100. 分析:dp[x][y]为从位置(x,y)出发的最长路. #pragma comment(linker, "/STACK:102400000, 102400000") #include<cstdio> #include<cstring> #include<cstdlib>…

这是一个简单的问题.你并不需要打印路径. 状态方程dp[i][j] = max(dp[i-1][j],dp[i][j-1],dp[i+1][j],dp[i][j+1]); 14003395 10285 option=com_onlinejudge&Itemid=8&page=show_problem&problem=1226" style="margin:0px; padding:0px; color:rgb(153,0,0); text-decoration:…

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…

[Link]: [Description] 在一个r*c的格子上; 求最长的下降路径; [Solution] 记忆化搜索; f[x][y]表示从(x,y)这个格子往下还能走多远; 因为是严格递增,所以有单调性. [NumberOf WA] 0 [Reviw] [Code] #include <bits/stdc++.h> using namespace std; const int N = 100+10; const int dx[5] = {0,1,-1,0,0}; const int dy…