题意:求一个序列中的最大 m 段和,m 段不能交叉. dp[i][0/1][j] 表示已经取完第 i 个物品,第 i 个物品取或不取,取到第 j 个子段. 用vis[i][0/1][j] 表示该 dp 值是否存在. 然后当 vis[i][0][j] 存在,即第 i 个物品不取,之前已经取了 j 个子段,可推得: 第 i+1 个不取: dp[i+1][0][j]=max(dp[i+1][0][j],dp[i][0][j]); 第 i+1 个取: dp[i+1][1][j+1]=max(dp[i+1…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 31583    Accepted Submission(s): 11174 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" proble…

/* dp[i][j]表示取第i个数时分成了j块 要么是将第i个数加入j块中的最后一块,要么是自成一块,加上前面j-1块的和 状态转移方程: dp[i][j]=max(dp[i-1][j]+a[i],max{dp[0][j-1]...dp[i-1][j-1]}) 枚举时j为外层循环,i为内层循环, 用滚动数组压缩j,再记录上一轮的dp[0..i][j]的最大值即可 */ #include<bits/stdc++.h> using namespace std; #define maxn 1000…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 27976    Accepted Submission(s): 9749 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem.…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 17164    Accepted Submission(s): 5651 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem…

Max Sum Plus Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 41885    Accepted Submission(s): 15095 Problem Description Now I think you have got an AC in Ignatius.L's "Max Sum" problem…

题目大意: 在n个数,求不重复的m段中的数据总和的最大值 令dp[i][j]表示将前j个数分成 i 段时得到的最大值(必取到第 j 个数) 状态转移可列为 dp[i][j]=Max(dp[i][j-1]+a[j] , Max( dp[i-1][k] ) + a[j] ) 0<k<j #include <cstdio> #include <cstring> #include <string> #include <iostream> using na…

链接:http://acm.hdu.edu.cn/showproblem.php?pid=1024 #include<iostream> #include<vector> #include<algorithm> #include<cmath> #include<cstring> #include<queue> #include<cstdio> #include<unordered_map> #define in…

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 consecutive number sequence S 1, S 2, S 3,…

http://acm.hdu.edu.cn/showproblem.php?pid=1024 最大m字段和,题目就不多说了,经典dp 这题坑爹...首先不说明m的范围(n<=1000000),还以为有O(n)的算法,吓得不敢做.其次,按照题目给的范围完全可能超int,但是数据事实上int远远够了,害得我__int64读比读int慢两倍,直接TLE.就是因为TLE的原因,害得我数组初始化没有考虑全(其实应该怪自己没有考虑清楚->_->),但事实上,完全可以每次memset一遍,因为超时,不…