Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23697 Accepted Submission(s): 8094
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
没做出来。。只能想到二维的，滚动数组果然厉害.
题意：n个数字分成m段的最大和
分析：dp[i][j] 代表以a[j]结尾的前j个数字被分成 i 段得到的最大和
可以得到: 1.如果a[j]单独成段 dp[i][j] = dp[i-1][k] + a[j] 其中 1<=k<j 意思是前1- k 个数字组成了i-1段.
2.如果a[j]并入第i段 那么dp[i][j]=dp[i][j-1]+a[j] 分析可得 dp[i][j] = max(1,2)
但是这题的条件是不允许这样做的 ，首先枚举 i , j ,k 的时间复杂度是 O(n^3) dp数组的空间要 O(n^2)，而数据量已经到达了 1000000 显然不允许.于是，这里就要用一个新的思想了:滚动数组.
我们可以看到dp[i][j]只和是否包含a[j]相关，所以这里我们可以用两个一维数组存当前状态与前一个状态.
新的状态: dp[j]表示以a[j]结尾的前i段的最大和,pre[j]表示前j个数组成前i段的最大和,不一定包括a[j]
dp[j] = max(dp[j-1]+a[j],pre[j-1]+a[j])
/**题意：n个数字分成m段的最大和*/ ///分析：dp[i][j] 代表以a[j]结尾的前j个数字被分成 i 段得到的最大和 ///可以得到: 1.如果a[j]单独成段 dp[i][j] = dp[i-1][k] + a[j] 其中 1<=k<j 意思是前1- k 个数字组成了i-1段. /// 2.如果a[j]并入第i段 那么dp[i][j]=dp[i][j-1]+a[j] 分析可得 dp[i][j] = max(1,2) ///但是这题的条件是不允许这样做的 ，首先枚举 i , j ,k 的时间复杂度是 O(n^3) dp数组的空间要 O(n^2)，而数据量已经 ///到达了 1000000 显然不允许.于是，这里就要用一个新的思想了:滚动数组. ///我们可以看到dp[i][j]只和是否包含a[j]相关，所以这里我们可以用两个一维数组存当前状态与前一个状态. ///新的状态: dp[j]表示以a[j]结尾的前i段的最大和,pre[j]表示前j个数组成前i段的最大和,不一定包括a[j] ///dp[j] = max(dp[j-1]+a[j],pre[j-1]+a[j]) #include<stdio.h> #include<iostream> #include<string.h> #include<math.h> #include<algorithm> using namespace std; const int N = 1000005; int a[N]; int dp[N]; int pre[N]; int main() { int m,n; while(scanf("%d%d",&m,&n)!=EOF){ for(int i=1;i<=n;i++){ scanf("%d",&a[i]); dp[i]=pre[i]=0; } dp[0]=pre[0]=0; int mam; for(int i=1;i<=m;i++) {///枚举每一段 mam = -0x7fffffff; for(int j=i;j<=n;j++){ dp[j] = max(dp[j-1]+a[j],pre[j-1]+a[j]); pre[j-1] = mam; ///表示前 j-1 个数组成i段能够表示的最大和 mam=max(dp[j],mam); } } printf("%d ",mam); } return 0; }