HDU——1024 Max Sum Plus Plus (dynamic programming + optimization)

Link to the original question: http://acm.hdu.edu.cn/showproblem.php?pid=1024

Test sample

Sample Input
1 3 1 2 3
2 6 -1 4 -2 3 -2 3
Sample Output
6
8

Meaning: give you a length of nninteger sequence of$n$$num$ , let you find$The$ maximum sum of$m$ non-repeated sub-segments.

Problem-solving ideas: This problem should be easy to think of dynamic programming, and the state is easier to find.We use $dp\left[i\right]\left[j\right]$ to express$i$ sub-segments are$num\left[j\right]$ The largest sub-segment sum at the end. Then our initial state must be$dp\left[0\right]\left[0\right]$ , the final state is$max(dp[m][t]$,$m\le t\le n$。So how do we write this state transition equation? Think about it, isn't it$dp[i][j]=max(dp[i-1][t],dp[i][j-1])+num[j],(i\le t\le j-1)$。(Because of our state definition, we must add the ending element, so the key is $max$ function, that is, the previous state is either the previous$i-1$ maximum plus this to$num\left[j\right]$ as a new sub-segment, or pre-iiAddnum [j] num[j] tothe last field of$i$ subsections$num\left[j\right]$ .) Knowing this, can we start to do the problem? Let's observe$n$ range, up to$1e6$ . We simply can't open such a large array. And we have to perform a lot of redundant search operations (because we find every time$dp\left[i\right]\left[j\right]$ , must find$dp[i-1]$ The maximum value of$[$$t$$]$ . ) This time complexity height$O(m\times n^2)$. This is obviously not feasible, so we need to optimize.

Optimization: Let’s take another look, for the two-dimensional dp dp we opened$dp$ array, can we find a way to open it into one dimension?Of course, we can use loops to replace this effect, that is, perform $m$ cycles, each cycle will add a new sub-segment. Then of course we want to optimize this new subsection.Let's look at the state transition equation again. According to our thinking, we need to solve and solve $dp[i-1]\; The$ problem of the maximum value of$[$$t$$]$ , is this we found that we have already sought it in the previous state? That is, at the$i-In\; 1$ cycle, the maximum value we solved. Of course we can store this value. Then we can introduce an array$pre\_max$ array, use$pre\_max[j-1]$ means$dp[i-1]$ The maximum value of$[$$t$$]$ , this maximum value is naturally easy to find, that is, it is solved in the previous cycle. OK, look at the code specifically.

AC code

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#include<bits/stdc++.h>	//POJ不支持

#define rep(i,a,n) for (int i=a;i<=n;i++)//i为循环变量，a为初始值，n为界限值，递增
#define per(i,a,n) for (int i=a;i>=n;i--)//i为循环变量， a为初始值，n为界限值，递减。
#define pb push_back
#define IOS ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
#define fi first
#define se second
#define mp make_pair

using namespace std;

const int inf = 0x3f3f3f3f;//无穷大
const int maxn = 1e6+4;//最大值。
typedef long long ll;
typedef long double ld;
typedef pair<ll, ll>  pll;
typedef pair<int, int> pii;
//*******************************分割线，以上为自定义代码模板***************************************//

int n,m;
int num[maxn];
int dp[maxn];
int pre_max[maxn];
int main(){
    
    
	//freopen("in.txt", "r", stdin);//提交的时候要注释掉
	IOS;
	while(cin>>m>>n){
    
    
		rep(i,1,n){
    
    
			cin>>num[i];
		}
		memset(dp,0,sizeof(dp));
		memset(pre_max,0,sizeof(pre_max));
		int temp;
		rep(i,1,m){
    
    
			//m层循环，每层循环添加一个字段。但注意，我们添加的字段只能从i到n中添加。原因是因为字段必须包含一个元素。
			temp=-inf;//先初始化为最小值，这个保存的是临时最大值。
			rep(j,i,n){
    
    
				//这里利用的特点就是取消了第一维，转化为i来代替。
				dp[j]=max(pre_max[j-1],dp[j-1])+num[j];
				pre_max[j-1]=temp;//pre_max[j-1]存放i到j-1中的最大值。
				temp=max(temp,dp[j]);//temp存放i到j的最大值。
			}
		}
		cout<<temp<<endl;
	}
	return 0;
}