Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17436 Accepted Submission(s): 5731
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. ^_^
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.
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）
题目大意：给你两个数M和N，之后是N个数，从这N个数找到M个子段，
求M个子段的最大和
思路：一开始不懂怎么找状态转移方程。参考别人博客才明白。
.设dp[i][j] 为将前 j 个数字分成 i 段的最大和。num[j]为当前数字
那么转移方程为 dp[i][j] = max(dp[i][j-1]+num[j]，dp[i-1][k]+num[j]) (i-1<=k<=j-1)
也可以视为 dp[i][j] = max(dp[i][j-1]+num[j]，max(dp[i-1][i-1],dp[i-1][i],…,dp[i-1][j-1])
)
//注：max(dp[i-1][i-1],dp[i-1][i],…,dp[i-1][j-1]) 就是dp[i-1][j-1]
意思是：前 j 个数字分成 i 段的最大和有两个决策。
1、将当前第j个数字并入第i段，与第j-1个数字所在的一段并为一段的最大和。
2、将当前第j个数字作为第i段，而第k个数字所在的一段为第i-1段，区间(k+1,j-1)的数字
不再选则的最大和。
取这两个决策中最大值。
本题还有一个难点在于将二维转为一位数组。考虑到第i行的状态由第i-1行和第i行递推过来，
所以可以利用滚动数组将二维压缩为一维数组，过程有点不太理解，留到以后慢慢想。
参考博文：http://blog.csdn.net/acm_davidcn/article/details/5887401
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int dp[1000010];
int maxn[1000010];
int num[1000010];
int main()
{
int M,N;
while(~scanf("%d%d",&M,&N))
{
dp[0] = maxn[0] = 0;
for(int i = 1; i <= N; i++)
{
scanf("%d",&num[i]);
dp[i] = maxn[i] = 0;
}
int MAXN;
for(int i = 1; i <= M; i++)//分为i段
{
MAXN = -0xffffff0;
for(int j = i; j <= N; j++)//第j个数字
{
dp[j] = max(dp[j-1]+num[j],maxn[j-1]+num[j]);
maxn[j-1] = MAXN;
MAXN = max(MAXN,dp[j]);
}
}
printf("%d\n",MAXN);
}
return 0;
}