Maximum sum——dp-优快云博客

描述

Given a set of n integers: A={a1, a2,…, an}, we define a function d(A) as below:
                t1     t2 
    d(A) = max{ ∑ai + ∑aj | 1 <= s1 <= t1 < s2 <= t2 <= n }
               i=s1   j=s2
Your task is to calculate d(A).

输入

The input consists of T(<=30) test cases. The number of test cases (T) is given in the first line of the input.
Each test case contains two lines. The first line is an integer n(2<=n<=50000). The second line contains n integers: a1, a2, …, an. (|ai| <= 10000).There is an empty line after each case.

输出

Print exactly one line for each test case. The line should contain the integer d(A).

样例

Input
1
10
1 -1 2 2 3 -3 4 -4 5 -5
Output
13

题解

题意就是给一个序列，找其中两个不相交的子串s1、s2，使得s1+s2最大
用dpl[i]表示以i为结尾的最大子串和，用dpr[i]表示以i为开头的最大子串和，再用一个dpres[i]表示在i后面（包括i）的最大子串和，即max(dpr[j] (i <= j < n))。
状态转移方程：
dpl[i]=max(dpl[i-1]+num[i],num[i]);
dpr[i]=max(dpr[i+1]+num[i],num[i]);
dpres[i]=max(dpres[i+1],dpr[i]);

Code

#include<bits/stdc++.h>
#define INIT(a,b) memset(a,b,sizeof(a))
#define LL long long int
using namespace std;
const int MAX=0x7fffffff;
const int MIN=-0x7fffffff;
const int INF=0x3f3f3f3f;
const int Mod=1e9+7;
const int MaxN=1e7+7;
int dpl[50005],dpr[50005],num[50005],dpres[50005];
int main(){
    //freopen("1.in","r",stdin);
    //freopen("1.out","w",stdout);
	int T;scanf("%d",&T);
	while(T--){
        INIT(dpl,0);INIT(dpr,0);INIT(dpres,0);
        int n;scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",&num[i]);

        dpl[1]=num[1],dpr[n]=dpres[n]=num[n];
        for(int i=2;i<=n;i++)
            dpl[i]=max(dpl[i-1]+num[i],num[i]);
        for(int i=n-1;i>=1;i--){
            dpr[i]=max(dpr[i+1]+num[i],num[i]);
            dpres[i]=max(dpres[i+1],dpr[i]);
        }

        int ans=MIN;
        for(int i=2;i<=n;i++)
            ans=max(dpres[i]+dpl[i-1],ans);
        printf("%d\n",ans);
	}
    return 0;
}