Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 35117 Accepted Submission(s): 12510
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 S 1, S 2, S 3, S 4 ... S x, ... S n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S x ≤ 32767). We define a function sum(i, j) = S i + ... + S j (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(i 1, j 1) + sum(i 2, j 2) + sum(i 3, j 3) + ... + sum(i m, j m) maximal (i x ≤ i y ≤ j x or i x ≤ j y ≤ j x 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(i x, j x)(1 ≤ x ≤ m) instead. ^_^
Given a consecutive number sequence S 1, S 2, S 3, S 4 ... S x, ... S n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S x ≤ 32767). We define a function sum(i, j) = S i + ... + S j (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(i 1, j 1) + sum(i 2, j 2) + sum(i 3, j 3) + ... + sum(i m, j m) maximal (i x ≤ i y ≤ j x or i x ≤ j y ≤ j x 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(i x, j x)(1 ≤ x ≤ m) instead. ^_^
Input
Each test case will begin with two integers m and n, followed by n integers S
1, S
2, S
3 ... S
n.
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.
题意:输入m,n,然后有n个数,求m段总和最大。
这道题前面想着把它分为m段,然后求这m段的最大子段和。后来越弄越糟,直接放弃这个思路。又想着暴力组合一下各个的和。想不出思路,就去找题解了。(萌新果然还是太嫩了qwq)
后面,单纯只看到这个状态转移方程都写不出能a的代码。。。。。
状态转移方程: dp[i][j]=max(dp[i][j-1]+num[j],dp(i-1,t)+num[j]) 其中i-1<=t<=j-1.
1 #include<cstdio> 2 using namespace std; 3 const int maxn=1000006; 4 int a[maxn],dp[maxn]; 5 6 int my_max(int x,int y) 7 { 8 return x>y?x:y; 9 } 10 11 int main() 12 { 13 int n,m; 14 while( ~scanf("%d%d",&m,&n)){ 15 16 for(int i=1;i<=n;i++){ 17 scanf("%d",&a[i]); 18 dp[i]=0; 19 } 20 21 int temp; 22 for(int i=1;i<=m;i++){ 23 temp=0; 24 for(int j=1;j<=i;j++) 25 temp+=a[j]; 26 dp[n]=temp; 27 28 for(int j=i+1;j<=n;j++){ 29 temp=my_max( dp[j-1], temp)+a[j]; 30 dp[j-1]=dp[n]; 31 dp[n]=my_max( dp[n], temp); 32 } 33 } 34 35 printf("%d\n",dp[n]); 36 } 37 return 0; 38 }
https://www.cnblogs.com/dongsheng/archive/2013/05/28/3104629.html