Super Jumping! Jumping! Jumping! 搬中文ing Descriptions: wsw成功的在zzq的帮助下获得了与小姐姐约会的机会,同时也不用担心wls会发现了,可是如何选择和哪些小姐姐约会呢?wsw希望自己可以循序渐进,同时希望挑战自己的极限,我们假定每个小姐姐有一个“攻略难度值” 从攻略成功第一个小姐姐开始,wsw希望每下一个需要攻略的小姐姐难度更高,同时又希望攻略难度值之和最大,好了,现在小姐姐们排成一排,wsw只能从左往右开始攻略,请你帮助他找到最大的攻略难…

http://acm.hdu.edu.cn/showproblem.php?pid=1087 设dp[i]表示去到这个位置时的最大和值.(就是以第i个为结尾的时候的最大值) 那么只要扫描一遍dp数组,就能得到ans,因为最后一步可以无条件到达终点. 那么可以用O(n^2)转移,枚举每一个位置,其中要初始化dp值,dp[i] = a[i],意思就是到达第i个位置的时候,和值最小也是 a[i]把,因为无论如何也可以一步到达a[i]这个值. #include <cstdio> #include &l…

Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 49436    Accepted Submission(s): 22871 Problem Description Nowadays, a kind of chess game called "Super Jumping!…

Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 56910    Accepted Submission(s): 26385 Problem Description Nowadays, a kind of chess game called “Super Jumping!…

HDU 1087 题目大意:给定一个序列,只能走比当前位置大的位置,不可回头,求能得到的和的最大值.(其实就是求最大上升(可不连续)子序列和) 解题思路:可以定义状态dp[i]表示以a[i]为结尾的上升子序列的和的最大值,那么便可以得到状态转移方程 dp[i] = max(dp[i], dp[j]+a[i]), 其中a[j]<a[i]且j<i; 另外每个dp[i]可以先初始化为a[i] 理解:以a[i]为结尾的上升子序列可以由前面比a[i]小的某个序列加上a[i]来取得,故此有dp[j]+a[…

http://acm.hdu.edu.cn/showproblem.php?pid=1087   Super Jumping! Jumping! Jumping! Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1087 Description Nowadays, a kind of chess game called “Super Ju…

Super Jumping! Jumping! Jumping! Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1087 Appoint description:  System Crawler  (2017-04-13) Description Nowadays, a kind of chess game called “Super…

Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 33384    Accepted Submission(s): 15093 Problem Description Nowadays, a kind of chess game called “Super Jumping!…

H - Super Jumping! Jumping! Jumping! Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1087 Appoint description:  System Crawler  (2015-11-18) Description Nowadays, a kind of chess game called “Su…

Super Jumping! Jumping! Jumping! Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1087 Appoint description:  System Crawler  (2015-09-05) Description Nowadays, a kind of chess game called “Super…

作者:jostree 转载请注明出处 http://www.cnblogs.com/jostree/p/4092939.html 题目链接:hdu 5025 Saving Tang Monk 状态压缩dp+广搜 使用dp[x][y][key][s]来记录孙悟空的坐标(x,y).当前获取到的钥匙key和打死的蛇s.由于钥匙具有先后顺序,因此在钥匙维度中只需开辟大小为10的长度来保存当前获取的最大编号的钥匙即可.蛇没有先后顺序,其中s的二进制的第i位等于1表示打死了该蛇,否则表示没打死.然后进行广度…

HDU 3016 Man Down (线段树+dp) Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1391    Accepted Submission(s): 483 Problem Description The Game “Man Down 100 floors” is an famous and interesting ga…

Lost's revenge Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 3757    Accepted Submission(s): 1020 Problem Description Lost and AekdyCoin are friends. They always play "number game"(A bor…

题意:给你几个模式串,问你主串最少改几个字符能够使主串不包含模式串 思路:从昨天中午开始研究,研究到现在终于看懂了.既然是多模匹配,我们是要用到AC自动机的.我们把主串放到AC自动机上跑,并保证不出现模式串,这里对AC自动机的创建有所改动,我们需要修改不存在但是符合要求的节点,如果某节点的某一子节点不存在,我们就把这个子节点指向他父辈节点存在的该节点(比如k->next[1]不存在,k->fail->next[1]存在,我们就把他改为k->next[1] = k->fail-…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1087 Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 24452    Accepted Submission(s): 10786 Problem Description No…

http://acm.hdu.edu.cn/showproblem.php?pid=1028 dp[i][j]表示数值为i,然后最小拆分的那个数是j的时候的总和. 1 = 1 2 = 1 + 1 .  2 = 2 3 = 1 + 1 + 1. 3 = 2 + 1. 3 = 3 那么可以分两类, 1.最小拆分数是j,这个时候dp[i][j] = dp[i - j][j].加一个数j,使得它变成i 2.最小拆分数严格大于j,这个时候就没得加上j了.就是dp[i][j + 1] 所以dp[i][j]…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1087 ---------------------------------------------------------------------------------------------------------------------------------------------------------- 欢迎光临天资小屋:http://user.qzone.qq.com/593830943…

Description Nowadays, a kind of chess game called "Super Jumping! Jumping! Jumping!" is very popular in HDU. Maybe you are a good boy, and know little about this game, so I introduce it to you now. The game can be played by two or more than two…

Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 34052    Accepted Submission(s): 15437 Problem Description Nowadays, a kind of chess game called “Super Jumping!…

C - Super Jumping! Jumping! Jumping! Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Description Nowadays, a kind of chess game called “Super Jumping! Jumping! Jumping!” is very popular in HDU. Maybe you are a good boy, a…

Problem Description Nowadays, a kind of chess game called "Super Jumping! Jumping! Jumping!" is very popular in HDU. Maybe you are a good boy, and know little about this game, so I introduce it to you now. The game can be played by two or more t…

解题思路:题目的大意是给出一列数,求这列数里面最长递增数列的和 dp[i]表示到达地点i的最大值,那么是如何达到i的呢,则我们可以考虑没有限制条件时候的跳跃,即可以从第1,2,3,---,i-1个地点跳跃到i, 而题目限定了,跳到的那个点的数要比开始跳的那个点的数大 所以,状态转移方程式为 for(i=1;i<=n;i++)   for(j=0;j<i;j++) if(a[j]>a[i])   dp[i]=max(dp[j]+a[i],dp[i]);//找出到达地点i的最大值 反思:本来…

Super Jumping! Jumping! Jumping!Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 47017    Accepted Submission(s): 21736 Problem DescriptionNowadays, a kind of chess game called “Super Jumping! Ju…

DP基础题 DP[i]表示以a[i]结尾所能得到的最大值 但是a[n-1]不一定是整个序列能得到的最大值 #include <bits/stdc++.h> using namespace std; ; int dp[maxn],n,a[maxn]; int main() { while(scanf("%d",&n)&&n) { memset(dp,,sizeof(dp)); ;i<n;i++) scanf("%d",&…

以下引用自:http://www.cnblogs.com/Lyush/archive/2011/08/31/2161314.html沐阳 该题可以算是一道经典的DP题了,题中数据是这样的.以 3 1 3 2 为例,首先 3 代表有三个数, 后面给出三个数,求该串的一个子串,使得其值一直是递增的,而且要求输出最大的和值.可以论证,该子串一定会是最长上升子串,因为,如果一个串还能够插入一个元素的话,那么这个串就一定不是最大的和了.而这个最长的上升子串还满足是所有同样长度的子串中最优的,和值最大. 具…

题目链接 DP基础题 求的是上升子序列的最大和 而不是最长上升子序列LIS DP[i]表示以a[i]结尾所能得到的最大值 但是a[n-1]不一定是整个序列能得到的最大值 #include <bits/stdc++.h> using namespace std; ; int dp[maxn],n,a[maxn]; int main() { while(scanf("%d",&n)&&n) { memset(dp,,sizeof(dp)); ;i<…

传送门:HDU_1087 题意:现在要玩一个跳棋类游戏,有棋盘和棋子.从棋子st开始,跳到棋子en结束.跳动棋子的规则是下一个落脚的棋子的号码必须要大于当前棋子的号码.st的号是所有棋子中最小的,en的号是所有棋子中最大的.最终所得分数是所有经过的棋子的号码的和. 思路:读完题之后知道这是一个最长上升子序列的题目.因为之前刚刚看过牛客网上一节讲解最长上升子序列的视屏,所以一上来就找准了方向,but我只知道怎么求最长上升子序列的长度啊,和怎么求???于是自己想方法开始求和,然后就wa掉了一个上午.…

题意:从起点依次跳跃带有数字的点直到终点,要求跳跃点上的数字严格递增,问跳跃点的最大数字和. 分析: 1.若之前的点比该点数字小,则可进行状态转移,dp[i] = max(dp[i], dp[j] + a[i]); 2.dp[i]---截止到i,跳跃的最大数字和. 3.由于不确定最终是哪个点直接跳往终点可保证数字和最大,因此,扫一遍,ans = max(ans, dp[i]); #include<cstdio> #include<cstring> #include<cstdl…

题意:给定一个长度为n的序列,让你求一个和最大递增序列. 析:一看,是不是很像LIS啊,这基本就是一样的,只不过改一下而已,d(i)表示前i个数中,最大的和并且是递增的, 如果 d(j) + a[i] > d(i)  d(i) = d(j) + a[i] (这是保证和最大),那么怎么是递增呢?很简单嘛,和LIS一样 再加上a[i] > a[j],这不就OK了. 代码如下: #include <cstdio> #include <iostream> #include &l…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1087 Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 47055    Accepted Submission(s): 21755 Problem Description N…