POJ 2479 Maximum sum【最大连续和2】

Description

Given a set of n integers: A={a1, a2,..., an}, we define a function d(A) as below: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

分析：

动态规划。枚举两个子串的分割点i,1<=i<=n,分别求两个串的最大子串，求出的最大值即为所求。
设dp[i]为以元素i结束的子串的最大值，则dp[i]=max{dp[i-1]+a[i],a[i]}。设lt[i]为串a[1]-a[i]的最大子串的值，

则lt[i]=max{lt[i-1],dp[i]}，同理可求rt[i]。
code：

 

#include<stdio.h>
#define max(A,B)((A)>=(B)?(A):(B))
const int INF=100005;
int a[INF];
int lt[INF];
int rt[INF];
int dp[INF];
int main()
{
 int i,j,c,n,sum;
 scanf("%d",&c);
 while(c--)
 {
  scanf("%d",&n);
  for(i=1;i<=n;i++)
   scanf("%d",&a[i]);
  dp[n+1]=dp[0]=-INF;
  lt[0]=rt[n+1]=-INF;

  //正向求。从左向右求dp[i]、lt[i]
  for(i=1;i<=n;i++)
   dp[i]=max(dp[i-1]+a[i],a[i]);
  for(i=1;i<=n;i++)
   lt[i]=max(dp[i],lt[i-1]);

//逆向求。从右向左求dp[i]、rt[i]
  for(i=n;i>=1;i--)
   dp[i]=max(dp[i+1]+a[i],a[i]);
  for(i=n;i>=1;i--)
   rt[i]=max(dp[i],rt[i+1]);

//枚举两个子串的分隔点i
  sum=-INF;
  for(i=1;i<=n;i++)
   sum=max(sum,lt[i]+rt[i+1]);
  printf("%d\n",sum);
 }
 return 0;
}