http://acm.hdu.edu.cn/showproblem.php?pid=1024
Max Sum Plus Plus
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 ≤ jxor 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 ≤ jxor 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.题意：将一串数字分成m段子序列，求最大可能达到的总和。
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<iostream> 5 using namespace std; 6 #define N 1000005 7 #define INF -1000000000 8 int num[N],dp[N],mmax[N]; 9 /* 10 状态DP[i][j] 11 前j个数可以分成i组的和的最大值 12 DP[i][j]=max(dp[i][j-1]+num[j], max( dp[i-1][k] )+num[j]),0<k<j; 13 dp[i][j-1]是把num[j]加入到前面的一个组， 14 max(dp[i-1][k])+num[j]是把num[j]重新分在另一个组里面， 15 而max(dp[i-1][k])是分成i-1个组时候的和的最大值 16 开一个mmax数组每次都可以保存分成i-1个组的时候使用前j-1个数时候最大值 17 时间复杂度O(mn) 18 */ 19 int main() 20 { 21 int n,m; 22 while(~scanf("%d%d",&m,&n)){ 23 for(int i=1;i<=n;i++){ 24 scanf("%d",num+i); 25 } 26 memset(dp,0,sizeof(dp)); 27 memset(mmax,0,sizeof(mmax)); 28 int ans; 29 for(int i=1;i<=m;i++){ 30 ans=INF; 31 for(int j=i;j<=n;j++){ 32 dp[j]=max(dp[j-1],mmax[j-1])+num[j]; 33 mmax[j-1]=ans; 34 ans=max(dp[j],ans); 35 } 36 } 37 cout<<ans<<endl; 38 } 39 return 0; 40 }
2016-06-21