hdu1024---Max Sum Plus Plus - c++编程基础

Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18190 Accepted Submission(s): 5955

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.

Author
JGShining（极光炫影）

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Statistic | Submit | Discuss | Note

数据好像很弱 O（n^2）能过

dp[i][j][0] 表示前i个数组成j段，且不选入第i个数
dp[i][j][1]表示前i个数组成j段，且选入第i个数

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

对于j=1的情况，就是最大子段和
数据有点大，用下滚动数组

/*************************************************************************
    > File Name: dp14.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年02月13日 星期五 20时39分01秒
 ************************************************************************/

#include  #include 
    
      #include 
     
       #include 
      
        #include 
       
         #include 
        
          #include 
         
           #include 
          
            #include 
           
             #include 
            
              #include 
             
               using namespace std; const double pi = acos(-1); const int inf = 0x3f3f3f3f; const double eps = 1e-15; typedef long long LL; typedef pair 
              
                PLL; const int N = 1000100; int dp[2][N][2]; int arr[N]; int g[N]; int main () { int m, n; while (~scanf("%d%d", &m, &n)) { memset (dp, -inf, sizeof(dp)); for (int i = 1; i <= n; ++i) { scanf("%d", &arr[i]); } dp[0][0][0] = 0; g[0] = 0; int tmp = -inf; for (int i = 1; i <= n; ++i) { if (g[i - 1] + arr[i] > arr[i]) { dp[i % 2][1][1] = g[i - 1] + arr[i]; g[i] = g[i - 1] + arr[i]; } else { dp[i % 2][1][1] = arr[i]; g[i] = arr[i]; } dp[i % 2][1][0] = tmp; tmp = max (tmp, g[i]); for (int j = 2; j <= m; ++j) { if (i < j) { break; } if (i - 1 >= j) { dp[i % 2][j][1] = max (dp[1 - (i % 2)][j - 1][0], max (dp[1 - (i % 2)][j][1], dp[1 - (i % 2)][j - 1][1])) + arr[i]; } else { dp[i % 2][j][1] = max (dp[1 - (i % 2)][j - 1][0], dp[1 - (i % 2)][j - 1][1]) + arr[i]; } if (i != j) { dp[i % 2][j][0] = max (dp[1 - (i % 2)][j][0], dp[1 - (i % 2)][j][1]); } else { dp[i % 2][j][0] = -inf; } } } printf("%d\n", max (dp[n % 2][m][0], dp[n % 2][m][1])); } return 0; }