HDU 1024 Max Sum Plus Plus(m个子段的最大子段和)
传送门:http://acm.hdu.edu.cn/showproblem.php?pid=1024
Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 35988 Accepted Submission(s): 12807
Problem 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 a consecutive number sequence S1, S2, S3, S4 ... Sx, ... Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx ≤ 32767). We define a function sum(i, j) = Si + ... + Sj (1 ≤ i ≤ j ≤ n).
Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i1, j1) + sum(i2, j2) + sum(i3, j3) + ... + sum(im, jm) maximal (ix ≤ iy ≤ jx or ix ≤ jy ≤ jx is not allowed).
But I`m lazy, I don't want to write a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(ix, jx)(1 ≤ x ≤ m) instead. ^_^
Given a consecutive number sequence S1, S2, S3, S4 ... Sx, ... Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx ≤ 32767). We define a function sum(i, j) = Si + ... + Sj (1 ≤ i ≤ j ≤ n).
Now given an integer m (m > 0), your task is to find m pairs of i and j which make sum(i1, j1) + sum(i2, j2) + sum(i3, j3) + ... + sum(im, jm) maximal (ix ≤ iy ≤ jx or ix ≤ jy ≤ jx is not allowed).
But I`m lazy, I don't want to write a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(ix, jx)(1 ≤ x ≤ m) instead. ^_^
Input
Each test case will begin with two integers m and n, followed by n integers S1, S2, S3 ... Sn.
Process to the end of file.
Process to the end of file.
Output
Output the maximal summation described above in one line.
Sample Input
1 3 1 2 3
2 6 -1 4 -2 3 -2 3
Sample Output
6
8
Hint
Huge input, scanf and dynamic programming is recommended.
Author
JGShining(极光炫影)
分析:
m个子段的最大子段和
子段不重合
第一次写这个问题,还不是很懂
我是参考的大佬的博客:
回来更新一波
今天训练赛又遇到这个问题了,说说我的理解
这个问题的dp做其实跟LIS和LCS的有一丁点的像的
外层n循环,i
内层 0到i-1循环 j
每第i次选择都是在前面的i个中(不包括i)选最大的加上当前a【i】的值
上面都是自己瞎逼逼,不好理解,看下面
分析:
最大m段字段和问题是由最大字段和问题演变来的
最大字段和问题大家应该都知道
最大字段和是把每个数据看成一个对象,从而组成一个子段
核心:那么最大m段字段和,就是把一个子段看成一个对象,再进行最大字段和的操作,从而组成m个最大子段,所以是对象变了(自己也是突然想到)
code:
#include<stdio.h> #include<iostream> #include<math.h> #include<string.h> #include<set> #include<map> #include<list> #include<algorithm> using namespace std; typedef long long LL; int mon1[13]= {0,31,28,31,30,31,30,31,31,30,31,30,31}; int mon2[13]= {0,31,29,31,30,31,30,31,31,30,31,30,31}; int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}}; #define INF 0x7fffffff//无穷大 #define max_v 1000010 int a[max_v]; int now[max_v];// now[j]:包含第j个元素的最大和 int pre[max_v];// pre[j]:前j个元素的最大和,不包括第j个元素 int main() { int n,m,maxx; while(~scanf("%d %d",&m,&n)) { for(int i=1;i<=n;i++) scanf("%d",&a[i]); memset(now,0,sizeof(now)); memset(pre,0,sizeof(pre)); for(int i=1;i<=m;i++)//m个子段 { maxx=-INF; for(int j=i;j<=n;j++) { now[j]=max(now[j-1]+a[j],pre[j-1]+a[j]); //now[j] 有两种来源,一种是直接在第i个子段后面加a[j],一种是a[j]单独成为一个子段 pre[j-1]=maxx;//更新pre使得pre[j-1]是前j-1个中的最大子段和 maxx=max(now[j],maxx); } } printf("%d\n",maxx); } return 0; }
心之所向,素履以往