标签：... cep BMI comm mission one font challenge sim
Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34541 Accepted Submission(s): 12341
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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#include <iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
#define MAXN 1100000
#define INF 0x3f3f3f3f
int dp[MAXN];
int maxn[MAXN];
int a[MAXN];
int main()
{
int n,m;
std::ios::sync_with_stdio(false);
while(cin>>m>>n){
for(int i=1;i<=n;i++){
cin>>a[i];
maxn[i]=0;
dp[i]=0;
}
dp[0]=0;
maxn[0]=0;
int maxx;
for(int i=1;i<=m;i++){
maxx=-INF;
for(int j=i;j<=n;j++){
dp[j]=max(dp[j-1]+a[j],maxn[j-1]+a[j]);
maxn[j-1]=maxx;
maxx=max(maxx,dp[j]);
}
}
cout<<maxx<<endl;
}
return 0;
}
HDU 1024 Max Sum Plus Plus（基础dp)
标签：... cep BMI comm mission one font challenge sim
原文地址：https://www.cnblogs.com/luowentao/p/8975477.html
