hdu 1024 最大m段不相交线段和

【hdu 1024 最大m段不相交线段和】的更多相关文章

hdu 1024 最大m段不相交线段和

题目传送门//res tp hdu 数据范围1e6,若是开二维会爆考虑用滚动数组优化 #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> #define rep(i,a,b) for(int i=(a);i<=(b);++i) #define per(i,a,b) for(int i = (a);i>=(b);--i) #define fo(i,a,…

HDU 1024 Max Sum Plus Plus （动态规划）

HDU 1024 Max Sum Plus Plus (动态规划) Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem. Given…

HDU 1024 max sum plus

A - Max Sum Plus Plus Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1024 Appoint description: Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a bra…

HDU 1024 Max Sum Plus Plus --- dp+滚动数组

HDU 1024 题目大意:给定m和n以及n个数,求n个数的m个连续子系列的最大值,要求子序列不想交. 解题思路:<1>动态规划,定义状态dp[i][j]表示序列前j个数的i段子序列的值,其中第i个子序列包括a[j], 则max(dp[m][k]),m<=k<=n 即为所求的结果 <2>初始状态: dp[i][0] = 0, dp[0][j] = 0; <3>状态转移: 决策:a[j]自己成为一个子段,还是接在前面一个子段的后面方程: a[j]直接接在前面…

Max Sum Plus Plus HDU - 1024

Max Sum Plus Plus HDU - 1024 Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult problem. Given a consecutiv…

怒刷DP之 HDU 1024

Max Sum Plus Plus Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1024 Appoint description: System Crawler (2015-09-05) Description Now I think you have got an AC in Ignatius.L's "Max Sum&…

hdu 1024（dp）

传送门:Max Sum Plus Plus 题意:从n个数中选出m段不相交的连续子段,求这个和最大. 分析:经典dp,dp[i][j][0]表示不取第i个数且前i个数分成j段达到的最优值,dp[i][j][1]表示取了第i个数且前i个数分成j段达到的最优值. 那么有: dp[i][j][0]=max(dp[i-1][j][0],dp[i-1][j][1]). dp[i][j][1]=max(dp[i-1][j-1][0]+a[i],max(dp[i-1][j-1][1],dp[i][j][1])…