题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5115 Dire Wolf Time Limit: 5000/5000 MS (Java/Others)Memory Limit: 512000/512000 K (Java/Others) 问题描述 Dire wolves, also known as Dark wolves, are extraordinarily large and powerful wolves. Many, if not…

Dire Wolf Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)Total Submission(s): 2221    Accepted Submission(s): 1284 Problem Description Dire wolves, also known as Dark wolves, are extraordinarily large and powerfu…

渐渐认识到区域赛更侧重的是思维及基本算法的灵活运用,而不是算法的量(仅个人见解),接下来要更多侧重思维训练了. 区间DP,dp[i][j]表示从i到j最终剩余第i 与第j只的最小伤害值,设置0与n+1两个虚拟的狼,状态转移方程如下: dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j] + b[i] + b[j]) #include<cstdio> #include<iostream> #include<cstdlib> #inclu…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5115 题目大意:有一些狼,从左到右排列,每只狼有一个伤害A,还有一个伤害B.杀死一只狼的时候,会受到这只狼的伤害A和这只狼两边的狼的伤害B的和.若两只狼之间的狼都被杀了,这两只狼也算相邻.求杀掉一排狼的最小代价.解题思路:设dp[i][j]为消灭编号从i到j只狼的代价,那么结果就是dp[1][n]枚举k作为最后一只被杀死的狼,此时会受到a[k]和b[i-1] b[j+1]的伤害 取最小的即可.可列出…

考虑先枚举所有的物品中最后拿走的,这样就分成了2个子问题,即先拿完左边的,再拿完右边的,最后拿选出的那个.令dp(i,j)表示拿完[i,j]所有物品的最小代价.你可能会说,我们拿[i,j]这一段物品的时候,i左边和j右边的第一个物品可能会不断变化,影响i和j的最终价格.其实是不会的,还是想想一开始说的整个序列枚举最后被拿走的,然后不断分割成子问题的过程,我们都是先拿走[i,j]中的所有元素,之后才会去动i-1或者j+1.所以可以澄清一下dp状态含义:dp(i,j)表示原序列中把[i,j]中的元素…

Dire Wolf Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)Total Submission(s): 3815    Accepted Submission(s): 2266 Problem Description Dire wolves, also known as Dark wolves, are extraordinarily large and powerfu…

Problem Description Teacher Mai has n numbers a1,a2,⋯,anand n−1 operators("+", "-" or "*")op1,op2,⋯,opn−1, which are arranged in the form a1 op1 a2 op2 a3 ⋯ an.He wants to erase numbers one by one. In i-th round, there are n+…

Problem Description The TV shows such as You Are the One has been very popular. In order to meet the need of boys who are still single, TJUT hold the show itself. The show is hold in the Small hall, so it attract a lot of boys and girls. Now there ar…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5568 题意: 求所有长度为k的严格升序子序列的个数. 题解: 令dp[i][k]表示以i结尾的长度为k的所有严格升序子序列的个数,则有状态转移:dp[i][k]+=dp[j][k-1](其中,arr[i]>arr[j],并且i>j): 最后答案为dp[1][k]+...dp[n][k]. 代码: import java.util.*; import java.math.*; public cla…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4597 #include <cstdio> #include <cstring> #include <iostream> #include <cmath> #include <algorithm> #include <queue> #include <vector> using namespace std; ; ; const…