HDU

本题是连续最大子序列和的加强版。

题目大意为 从一序列中取出若干段,使得这几段的和最大.

不得不说。。DP真的好难啊→ →

基本的连续最大子序列和代码为

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int a[1234];int main(){    int n;    while(~scanf("%d",&n))    {        int i;        for(i=0;i<=n-1;i++)        {            scanf("%d",&a[i]);        }        int sum=0;        int m=0;        for(i=0;i<=n-1;i++)        {            sum+=a[i];            m=max(m,sum);            if(sum<0)sum=0;        }        printf("%d\n",m);    }    return 0;}

下面是题面：

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, S₃, 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(i₁, 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 S₁, S ₂, S ₃ ... S _n.
Process to the end of file.

Output

Output the maximal summation described above in one line.

Sample Input

1 3 1 2 32 6 -1 4 -2 3 -2 3

Sample Output

68          

Hint

Huge input, scanf and dynamic programming is recommended.         

我们可以先写出状态转移方程式：

dp[i][j]=max{dp[i][j-1]+a[j],max{dp[i-1][t]}+a[j]}    i-1=<t<j-1但数组开不了这么大，而且三重循环肯定会超时。所以做一下改进。用dp[j]数组代表以j-1结尾i个子段的和,mm[j-1]数组代表以j-1结尾i-1个子段的和这样用两个数组来存就大大节省了空间然后就高高兴兴的AC啦！

#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int INF=0x3f3f3f3f;int a[1123456];int dp[1123456];int mm[1123456];int main(){    int n,m;    while(~scanf("%d%d",&m,&n))    {        int i,j;        for(i=1;i<=n;i++)        {            scanf("%d",&a[i]);            mm[i]=0;            dp[i]=0;        }        dp[0]=0;        mm[0]=0;        int ma;        for(i=1;i<=m;i++)        {            ma=-INF;            for(j=i;j<=n;j++)            {                dp[j]=max(dp[j-1]+a[j],mm[j-1]+a[j]);//dp[j-1]表示的是以j-1结尾i个子段的和..mm[j-1]表示前j-1个元素i-1个子段的和                mm[j-1]=ma;//保证a[i]是一个独立的子段                ma=max(ma,dp[j]);            }        }        printf("%d\n",ma);    }    return 0;}