HDU-1003 Max Sum（经典DP）

最新推荐文章于 2022-11-06 15:03:56 发布

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最新推荐文章于 2022-11-06 15:03:56 发布

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分类专栏： HDU 文章标签： dp

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本文介绍了一种求解整数序列中最大和子序列的算法，并提供了完整的C++实现代码。通过动态规划的方法，该算法能高效地找出序列中的最大和及其起始和终止位置。

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Problem Description
Given a sequence a[1],a[2],a[3]…a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.

Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).

Output
For each test case, you should output two lines. The first line is “Case #:”, # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.

Sample Input
2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5

Sample Output
Case 1:
14 1 4

Case 2:
7 1 6

定义一个数组sum，
sum[i]含义：所有以a[i]（i=3，表示以3结尾）为结尾的序列的序列和构成一个集合，此集合的最大值就是sum[i]。

如1,2,3.所有以a[3]=3为结尾的序列的序列和集合是{6,5,3},因而sum[3]=6.
sum的状态转移方程：sum[i] = max{sum[i-1]+a[i], a[i]}
ans必定是sum[0···(k-1)]之一。由于要记录起始位置和结束位置，引入s数组记录获得sum的序列的起始元素的位置，而由sum的定义，sum[i]的结束位置是i不用另外记录。

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <iomanip>
#include <string>
#include <set>
#include <map>
#include <queue> 
#include <stack>
#include <vector>
#include <algorithm>
#include <sstream> 
#include <cstring>
using namespace std;

int main()
{
	int t,t1;
	cin>>t;
	t1=0;
	while(t1<t)
	{
		int n,ans=1;
		cin>>n;
		int a[n+1],sum[n+1],s[n+1];	//s保存以i结尾的最大序列的首元素位置 
		for(int i=1;i<=n;i++){
			cin>>a[i];
		}
		sum[1]=a[1];
		s[1]=1; 	//初始化 
		
		for(int i=2;i<=n;i++){
			if(sum[i-1]>=0){
				sum[i]=sum[i-1]+a[i];
				s[i]=s[i-1];
			}
			else{
				sum[i]=a[i];
				s[i]=i;
			}
			if(sum[ans]<sum[i]){
				ans=i;
			}
		}
		
		cout<<"Case "<<t1+1<<":\n"<<sum[ans]<<" "<<s[ans]<<" "<<ans<<"\n";
		t1++;
		if(t1!=t)
			cout<<endl; 
	}
	return 0;
}