Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 34644 Accepted Submission(s): 12376
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 S 1, S 2, S 3, S 4 ... S x, ... S n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S x ≤ 32767). We define a function sum(i, j) = S i + ... + S j (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(i 1, j 1) + sum(i 2, j 2) + sum(i 3, j 3) + ... + sum(i m, j m) maximal (i x ≤ i y ≤ j x or i x ≤ j y ≤ j x 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(i x, j x)(1 ≤ x ≤ m) instead. ^_^
Given a consecutive number sequence S 1, S 2, S 3, S 4 ... S x, ... S n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S x ≤ 32767). We define a function sum(i, j) = S i + ... + S j (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(i 1, j 1) + sum(i 2, j 2) + sum(i 3, j 3) + ... + sum(i m, j m) maximal (i x ≤ i y ≤ j x or i x ≤ j y ≤ j x 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(i x, j x)(1 ≤ x ≤ m) instead. ^_^
Input
Each test case will begin with two integers m and n, followed by n integers S
1, S
2, S
3 ... S
n.
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）
Recommend
题目大意：给你n个数，划分成m块，求m块连续串的和最大值
状态转移方程： dp[i][j]=max(dp[i][j-1]+a[j],max(dp[i-1][k])+a[j]);
i表示划分成i个区块，j表示前j个数，a[j]为答案的最后一个元素，必定包含。
两种情况
a[j]属于第i块，接在j-1后面；
a[j]不属于i块，独立成一块，再加上前k个元素分成i-1块的最大值，其中 0<k<j
两种情况求最大值，两点原因必须优化：1.N=1000000 ，数组开不到那么大；2.两个max,套三层循环，超时
观察状态转移方程，可以改写成滚动数组形式：dp[i][j]的决策只取决于前面j-1个元素
对于max(dp[i-1][k])这层循环可以用数组记录每次的最大值
#include <iostream>
#include <cstring>#include <algorithm>
using namespace std;
#define N 1000000
#define INF 0x7fffffff
int dp[N+10],Max[N+10];
int a[N+10];
int main()
{
int n,m,mmax;
while(scanf("%d %d",&m,&n)!=EOF)
{
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
memset(dp,0,sizeof(dp));
memset(Max,0,sizeof(Max));
for(int i=1;i<=m;i++) //分m块
{
mmax=-INF;
for(int j=i;j<=n;j++)//至少一个元素一块吧
{
dp[j]=max(Max[j-1],dp[j-1])+a[j];
Max[j-1]=mmax;
mmax=max(dp[j],mmax);
}
}
printf("%d\n",mmax);
}
return 0;
}