hdu1024HDU 1024 Max Sum Plus Plus（动态规划 很详很详解）
Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6725 Accepted Submission(s): 2251
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.
昨天和舍友BR聊到4点左右，看来放假了也得早点回去了（避免落单在宿舍）
言归正传，卡了那么久，DP独立思考
#include//1000ms卡到984ms ^^ 故意的哈(几个月以前的记录，这句废话不打算删) #include #include using namespace std; int dp[1222222],alone[1222222],a[1222222]; int main() { int i,j,n,m; while(~scanf("%d",&m)) { scanf("%d",&n); memset(dp,0,sizeof(dp)); memset(alone ,0,sizeof(alone)); for(i=1;i<=n;i++)scanf("%d",&a[i]); int tmax; //★循环i内的alone[j]:保存前j个数(含j)分i段时的最大和★ for(i=1;i<=m;i++)//★分i段 { tmax=-(1<<30); //printf("a[],alo[],dp[]\n"); for(j=i;j<=n;j++)//从i开始枚举:至少分i段,这里就直接前i个数i段 { dp[j]=_cpp_max(dp[j-1],alone[j-1])+a[j]; /*★★dp前者表示a[j]与a[j-1]所在的一段合并成一段。al后者表示以a[j]为首开始第 i 段 ★循环i内的alone[j]:保存前j个数(含j)分i段时的最大和★所以可以有alone[j-1]+a[j] alone在同一次的j循环总是走不到一起,能走到一起的是下个i的"j循环"//*///\ printf("%2d %2d %2d\n",a[j],alone[j-1],dp[j]); if(j>i)alone[j-1]=tmax; //★循环i内的alone[j]:保存前j个数(含j)分i段时的最大和(所以答案是最后一次循环(分n段)时候的tmax)★//\ alone巧妙的保存赋值到i层j-1而又不影响当前i-1层的j-1的使用，代表前j-1个分i段最大和.//\ 细心点会发现最大只保存到alone[n-1],reason:分i段时,枚举用的是分★i-1段层的alone[i-1至n-1](所以答案是最后一次循环时候( 可含a[n])的tmax)//\ 比如分n-1段,最后第1次进内循环时起点是j=n-1,即前面n-2个数单独分成n-2段(所以答案是最后一次循环(分m段)时候的tmax) if(tmax
/*
带★(写完解释的那天晚上想说，老人都看懂了......现在又+了些解释--发现不够精简!)
虽然表达能力很菜，不过已经讲得很清楚了，还不懂的话用下面的样例去运行观察吧（m改成6吧）
2 6
-1 4 -2 3 -2 3
取4再区3,-2,3 or 4,-2,3+3=8
*/