POJ 2479 Maximum Sum (DP)

题意：给定n个数，求两段连续子列的最大和。

思路：先从左向右dp，求出一段连续子列的最大和，再从右向左dp，求出两段连续子列的最大和，方法还是挺经典的。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <memory.h>
#include <cmath>
#include <bitset>
#include <queue>
#include <vector>
using namespace std;

const int MAXN = 100400;
const int INF = 0x4ffffff;
#define CLR(x,y) memset(x,y,sizeof(x))
#define ADD(x) x=((x+1)&BORDER)
#define IN(x) scanf("%d",&x)
#define OUT(x) printf("%d\n",x)
#define MIN(m,v) (m)<(v)?(m):(v)
#define MAX(m,v) (m)>(v)?(m):(v)
#define ABS(x) ((x)>0?(x):-(x))

int arr[MAXN],dp[MAXN];
int sum,n,ans;
int init()
{
    sum = 0;
    ans = -INF;
    CLR(dp,0);
    return 0;
}
int input()
{
    for(int i = 0; i < n; ++i)
        IN(arr[i]);
    return 0;
}
int work()
{
    int i,j,tmp;
    tmp = -INF;
    for(i = 0; i < n; ++i)
    {
        sum += arr[i];
        tmp = MAX(tmp,sum);
        dp[i] = tmp;
        sum = MAX(sum,0);
    }
    sum = 0;
    tmp = -INF;
    for(i = n-1; i > 0; --i)
    {
        sum += arr[i];
        tmp = MAX(sum,tmp);
        sum = MAX(sum,0);
        ans = MAX(tmp+dp[i-1],ans);
    }
    OUT(ans);
    return 0;
}
int main()
{
    int tt;
    while(IN(n))
    {
        if(!n)
            break;
        init();
        input();
        work();
    }
    return 0;
}