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. ^_^
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.
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\n",tmaxn); 41 } 42 return 0; 43 }
