hdu 1024 Max Sum Plus Plus (子段和最大问题)

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Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17336    Accepted Submission(s): 5701


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.


给定一个数组,求M段连续的子段和最大。dp[i][j]表示前i段选择第j个元素的最优解。

dp[i][j]=max(dp[i][j-1]+a[j] , max( dp[i-1][k] ) + a[j] ) 0<k<j

由于k<j,所以我们能够用两个一维数组,一个dp[i]记录当前行状态。一个d[i]记录下一行可选的最大值。

用64位会超时。但int能水过。。

</pre><pre name="code" class="cpp">/*
dp[i][j]=max(dp[i][j-1]+a[j] , max( dp[i-1][k] ) + a[j] ) 0<k<j
第j个元素选i段区间,对于当前元素a[j],能够把它接到第i段的第j-1位置构成一段。
或者单独成一段。

观察这个方程。对于dp[i][j-1]这一项,它和dp[i][j]同i,就是在一段。 用一维数组就能记录; dp[i-1][k]是i-1段j位置前能取的最大值。我们能够在计算当前第i段时, 把dp[i-1][k]记录下来,取最大值就好了。 */ #include<stdio.h> #include<string.h> #include<stdlib.h> #include<algorithm> using namespace std; #define ll __int64 const int inf=0x7fffffff; #define N 1000010 int a[N]; int pre[N]; //即dp[i-1][k]。记录j位置前可选值 int dp[N]; int main() { int i,j,n,m; int tmp,ans; while(~scanf("%d%d",&m,&n)) { for(i=1;i<=n;i++) { scanf("%d",&a[i]); dp[i]=0; pre[i]=0; } pre[0]=dp[0]=0; ans=-inf; for(i=1;i<=m;i++) { tmp=-inf; for(j=i;j<=n;j++) { dp[j]=max(dp[j-1]+a[j],pre[j-1]+a[j]); pre[j-1]=tmp; //tmp记录的是当前段j位置的最大值, tmp=max(tmp,dp[j]);//而数组pre[]是记录j-1位置的,所以要先取tmp,再更新tmp } } for(i=m;i<=n;i++) ans=max(ans,dp[i]); printf("%d\n",ans); } return 0; }






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