摘要：完全背包。+实现俩个大数相加。http://poj.org/problem?id=3181Dollar DayzTime Limit:1000MSMemory Limit:65536KTotal Submissions:3431Accepted:1356DescriptionFarmer John goes to Dollar Days at The Cow Store and discovers an unlimited number of tools on sale. During his first visit, the tools are selling variously for $ 阅读全文

摘要：多重背包。。。。。。悼念512汶川大地震遇难同胞——珍惜现在，感恩生活Time Limit: 1000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 12523Accepted Submission(s): 5292 Problem Description急！灾区的食物依然短缺！ 为了挽救灾区同胞的生命，心系灾区同胞的你准备自己采购一些粮食支援灾区，现在假设你一共有资金n元，而市场有m种大米，每种大米都是袋装产品，其价格不等，并且只能整袋购买。 请问：你用有限的资金最多能采 阅读全文

摘要：http://acm.hdu.edu.cn/showproblem.php?pid=260201背包：用二维数组实现。c[n][m]表示n种物品，背包容量为m的最大价值。状态方程为:f(n,m)=max{f(n-1,m), f(n-1,m-w[n])+P(n,m)}这就是书本上写的动态规划方程. Bone CollectorTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 24513Accepted Submission(s): 9911 Pr 阅读全文

摘要：http://acm.hdu.edu.cn/showproblem.php?pid=15051506的加强，从一维变二维。 City GameTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3959Accepted Submission(s): 1651 Problem DescriptionBob is a strategy game programming specialist. In his new city building game 阅读全文

摘要：对于每一块木板,Area=height[i]*(j-k+1)其中,j=height[i];找j,k成为关键,一般方法肯定超时,利用动态规划,如果它左边高度大于等于它本身,那么它左边的左边界一定满足这个性质,再从这个边界的左边迭代下去 http://acm.hdu.edu.cn/showproblem.php?pid=1506 Largest Rectangle in a HistogramTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 95. 阅读全文

摘要：http://acm.hdu.edu.cn/showproblem.php?pid=1231之前用并查集来做的；现在用dp来做：dp的状态方程：dp[i]=max(dp[i-1]+a[i],a[i]); 最大连续子序列Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 16818Accepted Submission(s): 7386 Problem Description给定K个整数的序列{ N1, N2, ..., NK }，其任意连续子序列 阅读全文

摘要：,0-1背包思路就是如果报销n张就必须报销n-1张。j表示可以报销的张数。状态方程：dp[j] = Max(dp[j], dp[j-1]+v[i]);//状态方程恶心地方:有这样的输入数据3 A:100 A:200 A:300http://acm.hdu.edu.cn/showproblem.php?pid=1864这题和最大连续子序列http://acm.hdu.edu.cn/showproblem.php?pid=1003一样的思路。 最大报销额Time Limit: 1000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/O 阅读全文

摘要：01背包的概率问题 http://acm.hdu.edu.cn/showproblem.php?pid=2955 状态方程：dp[j]=max(dp[j],dp[j-m[i]]*1-pi[i]);思路是：当前的概率基于前一种状态的概率，即偷n家银行而不被抓的概率等于偷n-1家银行不被转的概率乘以偷第n家银行不被抓的概率。用dp[i]表示偷价值为i时不被抓的概率，则状态转移方程为： dp[j]=max(dp[j],dp[j-m[i]]*(1-p[i])); RobberiesTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32. 阅读全文