随笔分类 -  dp

摘要:Hamburger Magi Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 437 Accepted Submission(s): 144 Pr 阅读全文
posted @ 2016-01-28 22:21 sCjTyC 阅读(412) 评论(0) 推荐(0)
摘要:Doing Homework Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6967 Accepted Submission(s): 3043 阅读全文
posted @ 2016-01-27 22:45 sCjTyC 阅读(938) 评论(2) 推荐(1)
摘要:Fibonacci StringTime Limit: 3000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5008Accepted Submission(s): 1690Pr... 阅读全文
posted @ 2016-01-23 12:36 sCjTyC 阅读(323) 评论(0) 推荐(0)
摘要:机器人的舞蹈Time Limit: 1000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 292Accepted Submission(s): 154Problem Descri... 阅读全文
posted @ 2016-01-20 15:57 sCjTyC 阅读(243) 评论(0) 推荐(0)
摘要:CRB and His BirthdayTime Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1112Accepted Submission(s): 55... 阅读全文
posted @ 2016-01-20 12:46 sCjTyC 阅读(209) 评论(0) 推荐(0)
摘要:http://acm.hdu.edu.cn/showproblem.php?pid=1421;思路:先将所给的椅子的价值按升序排列,举个例子,四张椅子的价值分别为a,b,c,d(a 2 #include 3 #include 4 #include 5 #include 6 #include 7 co... 阅读全文
posted @ 2015-12-03 10:19 sCjTyC 阅读(226) 评论(0) 推荐(0)
摘要:sequence2Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 677Accepted Submission(s): 244Problem Des... 阅读全文
posted @ 2015-11-30 22:54 sCjTyC 阅读(271) 评论(0) 推荐(0)
摘要:http://acm.hdu.edu.cn/showproblem.php?pid=5569;题目意思:从(1,1)点出发只能向右和向下走,到达(n,n)点时所得到的价值最小,价值是Let the numbers you go through become an arraya1,a2,...,a2k... 阅读全文
posted @ 2015-11-26 13:03 sCjTyC 阅读(195) 评论(0) 推荐(0)
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posted @ 2015-11-22 23:15 sCjTyC 阅读(8) 评论(0) 推荐(0)
摘要:题目和南阳那道题一样链接http://www.cnblogs.com/zzuli2sjy/p/4943774.html代码: 1 #include 2 #include 3 #include 4 #include 5 #include 6 #include 7 typedef long long l... 阅读全文
posted @ 2015-11-19 12:23 sCjTyC 阅读(313) 评论(0) 推荐(0)
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posted @ 2015-11-11 18:51 sCjTyC 阅读(8) 评论(0) 推荐(0)
摘要:这题是区间DP。看见dp就成傻逼了。题目的意思是ai表示每头狼的本身具备的攻击力,bi表示能提供给相邻狼的攻击力,以什么样的顺序去杀狼可以使狼对人的攻击最少。且如果狼被杀了,就不能为相邻的狼提供攻击力了。动态转移方程为dp[i][j]=min(dp[i][k]+dp[k][j]+a[k]+a[i]+... 阅读全文
posted @ 2015-11-11 13:10 sCjTyC 阅读(314) 评论(0) 推荐(0)
摘要:当时比赛时超时了,那时没学过树状数组,也不知道啥叫离散化(貌似好像现在也不懂)。百度百科——离散化,把无限空间中无限的个体映射到有限的空间中去,以此提高算法的时空效率。 这道题是dp题,离散化和树状数组用来优化,状态转移方程:dp[i][j]=sum(dp[i-1][k]) k需要满足a[j]>a[ 阅读全文
posted @ 2015-11-06 22:09 sCjTyC 阅读(848) 评论(0) 推荐(0)
摘要:刚写这题时知道是dp但动态转移方程推不出来,后来上网上搜了题解。先把问题简化,假如就选一段,这就是个经典的max sum dp。状态转移方程为dp[i]=dp[i-1]+a[i]>a[i]?:dp[i-1]+a[i]:a[i];题目:http://acm.hdu.edu.cn/showproblem... 阅读全文
posted @ 2015-10-30 10:44 sCjTyC 阅读(233) 评论(0) 推荐(0)