Given a set of n integers: A={a1, a2,..., an}, we define a function d(A) as below:
POJ 2479 Maximum sum (双向DP)
Your task is to calculate d(A).
Input
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.
Output
Print exactly one line for each test case. The line should contain the integer d(A).
Sample Input
1
10
1 -1 2 2 3 -3 4 -4 5 -5
Sample Output
13
Hint
In the sample, we choose {2,2,3,-3,4} and {5}, then we can get the answer.
Huge input,scanf is recommended.
POJ Contest,Author:Mathematica@ZSU
双向DP求解:
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int N=50010;
int n,num[N];
int dpl[N],dpr[N];
int main(){
//freopen("input.txt","r",stdin);
int t;
scanf("%d",&t);
while(t--){
scanf("%d",&n);
scanf("%d",&num[0]);
dpl[0]=num[0];
for(int i=1;i<n;i++){
scanf("%d",&num[i]);
if(dpl[i-1]>=0) //从左向右求最大值
dpl[i]=dpl[i-1]+num[i];
else
dpl[i]=num[i];
}
int maxv=num[n-1],ans=-999999;
dpr[n-1]=num[n-1];
for(int i=n-1;i>=1;i--){
if(dpr[i]>=0) //从右向左求最大值
dpr[i-1]=dpr[i]+num[i-1];
else
dpr[i-1]=num[i-1];
if(maxv<dpr[i])
maxv=dpr[i];
if(ans<maxv+dpl[i-1])
ans=maxv+dpl[i-1];
}
printf("%d\n",ans);
}
return 0;
}