hdu 1024 Max Sum Plus Plus(DP最大字段和)
Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11158 Accepted Submission(s): 3673
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. ^_^
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.
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 8HintHuge input, scanf and dynamic programming is recommended.
Author
JGShining(极光炫影)
思路:s[i][j]为i段包含j的最大和。ma[i][j]为前j个分为i段的最大和。答案就是ma[m][n]
d[j]存的是数
s[i][j]=max(s[i][j-1]+d[j],ma[i-1][j-1]+d[j]);
ma[i][j]=max(s[i][k]) (i<=k<=j)
又因为题目数给的太大一百万。所以二维数组不好开。想想可以用一维。
s[j]为i段含j最大和,ma【j】为i段前j个最大和。
s[j]=max(s[j-1]+d[j],ma[j-1]+d[j])
ma[j]=max(s[k]) (i<=k<=j)
ma[j]要先用后更新。这样就不会有问题了。
#include<cstdio> #include<cstring> const int mm=1000010; const int oo=1e9; int d[mm],s[mm],ma[mm]; int n,m; int main() { while(scanf("%d%d",&m,&n)!=EOF) { for(int i=1;i<=n;i++) scanf("%d",&d[i]); memset(s,0,sizeof(s)); memset(ma,0,sizeof(ma)); int _min,_max; for(int i=1;i<=m;i++) { _max=-oo;int j; for(j=i;j<=n;j++) {if(s[j-1]<ma[j-1]) s[j]=ma[j-1]+d[j]; else s[j]=s[j-1]+d[j]; ma[j-1]=_max; if(s[j]>_max) _max=s[j]; } ma[j-1]=_max; } printf("%d\n",_max); } }
The article write by nealgavin