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): 21363 Accepted Submission(s): 7144
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
http://www.cnblogs.com/lishuhuakai/archive/2012/10/13/4840323.html (感觉写的很不错, 反正我是看懂了)
这道题在国赛前我就看了, 可是一直不知如何下手, 终于决定一定要把它给A了, 现在的我能沉下心来, 好好看解析, 好好的自己去理解, 果然还是懂了的
重点在递归式上
w[i][j]: 前 j 个数分为 i 段, 第 j 个数必须选;1. 第 j 个数单独为1段;2. 第 j 个数与前面的数连一块。
w[i][j] = max(b[i-1][j-1], w[i][j-1]) + a[j];
b[i][j]:前 j 个数分为 i 段, 第 j 个数可选可不选; 1.选第 j 个数;2.不选第 j 个数。
b[i][j] = max(b[i][j-1], w[i][k]);
/// w[j] 表示 j 个元素取 i 段, a[j] 必须取是的最大值
w[j] = max(dp[1-t][j-1], w[j-1]) + sum[j]-sum[j-1];
/// dp[t][j] 表示在a[j]可取可不取这两种情况下取得的最大值
dp[t][j] = max(dp[t][j-1], w[j]);
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<iostream> #include<stdio.h> #include<string.h> #include<algorithm> #define N 1000001 using namespace std; int sum[N], w[N], dp[2][N]; ///sum[i] 里面存的是前 i 项和 int main() { int m, n; while(scanf("%d%d", &m, &n)!=EOF) { int i, j, x; sum[0] = 0; for(i=1; i<=n; i++) { scanf("%d", &x); sum[i] = sum[i-1] + x; dp[0][i] = 0; ///从前 i 个元素中取 0 段, 最大值为 0 } /** 我们先假设a[i]中 存放该序列的第 i 个值, w[i][j] 表示前 j 个数分为 i 段, 第 j 个数必须选这种情况下取得的最大值 b[i][j]表示在前 j 个数中取 i 段 这种情况写取得的最大值 w[i][j]: 前 j 个数分为 i 段, 第 j 个数必须选;1. 第 j 个数单独为1段;2. 第 j 个数与前面的数连一块。 w[i][j] = max(b[i-1][j-1], w[i][j-1]) + a[j]; b[i][j]:前 j 个数分为 i 段, 第 j 个数可选可不选; 1.选第 j 个数;2.不选第 j 个数。 b[i][j] = max(b[i][j-1], w[i][j]); **/ int t=1; for(i=1; i<=m; i++) /// i表示取 i 段 { for(j=i; j<=n; j++) /// 如果dp[i][j](j<i)是没有意义的 { if(i==j) dp[t][j] = w[j] = sum[j]; else { /// w[j] 表示 j 个元素取 i 段, a[j] 必须取是的最大值 w[j] = max(dp[1-t][j-1], w[j-1]) + sum[j]-sum[j-1]; /// dp[t][j] 表示在a[j]可取可不取这两种情况下取得的最大值 dp[t][j] = max(dp[t][j-1], w[j]); } } t = 1-t; ///t在 0 和 1 直间交替变换 /** 为什么要交换呢??? 这是为了要节省空间 仔细观察递归式 w[i][j] = max(b[i-1][j-1], w[i][j-1]) + a[j]; b[i][j] = max(b[i][j-1], w[i][j]); 我们发现,对于取 i 段, w[i][j] 只与 b[i-1][k-1] 和 w[i][k-1] 有关, 与之前的那一些项没有关系 因此我们的数组可以开小一点, 用更新来覆盖掉前面的值!!! **/ } printf("%d ", dp[m%2][n]); } return 0; }