HDU 1024 Max Sum Plus Plus 动态规划

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题目链接http://acm.hdu.edu.cn/showproblem.php?pid=1024

题目大意:n个数分成两两不相交的m段,求使这m段和的最大值。

解题思路:比较坑的点:n能过;long long超时,int AC。 

dp[i][j]:= 在选择第i个数的情况下前i个数分成j段的最大值
dp[i][j] = max(dp[i - 1][j] + a[i], max(dp[x][j - 1] -> dp[x][j - 1]) + a[i]) x < i

由于n<1000000,并且更新dp[i][j]时只用到了j和j-1的部分,因此采用滚动数组记录选择第i个人时前i个人分成j段的和最大值;同时,max(dp[x][j - 1] -> dp[x][j - 1]) 可以在更新dp[i]的同时记录,这样就将时间复杂度降低到了n2 (所以为什么n2能过???)

另外就是一些细节问题。

代码:

 1 const int inf = 0x3f3f3f3f;
 2 //dp[i][j]:= 在选择第i个数的情况下前i个数分成j段的最大值
 3 //dp[i][j] = max(dp[i - 1][j] + a[i], max(dp[x][j - 1] -> dp[x][j - 1]) + a[i]) x < i
 4 const int maxn = 1e6 + 5;
 5 int dp[maxn];
 6 int pre[maxn];
 7 int a[maxn], n, m;
 8 
 9 int solve(){
10     memset(dp, 0, sizeof(dp));
11     memset(pre, 0, sizeof(pre));
12     int tmax = -inf;
13     for(int j = 1; j <= m; j++){
14         tmax = -inf;//记录前i个人本次的最大值 
15         for(int i = j; i <= n; i++){
16             dp[i] = max(dp[i - 1] + a[i], pre[i - 1] + a[i]);//使用的是未更新的值,对应j-1的 
17             pre[i - 1] = tmax;    //再更新 
18             tmax = max(tmax, dp[i]);
19         }
20     }
21     return tmax; //不一定是dp[n],最后一个可能不用选 
22 }
23 
24 int main(){
25     while(scanf("%d %d", &m, &n) != EOF){
26         for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
27         int ans = solve();
28         printf("%d\n", ans);
29     }
30 }

题目:

Max Sum Plus Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 30900    Accepted Submission(s): 10889


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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