• hdu 1024 Max Sum Plus Plus (DP)


    Max Sum Plus Plus

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 15817    Accepted Submission(s): 5140


    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. ^_^
     
    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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     1 //609MS    8460K    692 B    C++
     2 /*
     3 
     4     求最大m段不重叠子序列和。
     5     
     6     DP:
     7     状态转移方程:
     8         dp[i][j]=max(dp[i][j-1]+a[j],max(dp[i-1][k])+a[j]) 0<k<j
     9         dp[i][j]表示i段时遍历到j并使用a[j]时最大值
    10         
    11     使用滚动数组 maxn[] 记录 max(dp[i-1][k]) 
    12 
    13 */
    14 #include<stdio.h>
    15 #include<string.h>
    16 #define N 1000005
    17 #define inf 0x7fffffff
    18 int dp[N],maxn[N],a[N];
    19 inline int Max(int x,int y)
    20 {
    21     return x>y?x:y;
    22 }
    23 int main(void)
    24 {
    25     int n,m;
    26     while(scanf("%d%d",&m,&n)!=EOF)
    27     {
    28         int tmaxn;
    29         memset(dp,0,sizeof(dp));
    30         memset(maxn,0,sizeof(maxn));
    31         for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    32         for(int i=1;i<=m;i++){
    33             tmaxn=-inf;
    34             for(int j=i;j<=n;j++){
    35                 dp[j]=Max(dp[j-1],maxn[j-1])+a[j];
    36                 maxn[j-1]=tmaxn;
    37                 tmaxn=Max(tmaxn,dp[j]);        
    38             }
    39         }
    40         printf("%d
    ",tmaxn);
    41     }
    42     return 0;
    43 }
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  • 原文地址:https://www.cnblogs.com/GO-NO-1/p/3744914.html
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