poj2593

dp，分别从左侧和右侧求最大子段和，然后把fleft[i]变成0～i区间中最大子段和，fright以此类推。

然后枚举中间分割点。

View Code

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;

#define maxn 100003

int f[maxn], fleft[maxn], fright[maxn];
int n;

void input()
{
for (int i = 0; i < n; i++)
scanf("%d", &f[i]);
}

void work1()
{
fleft[0] =f[0];
for (int i = 1; i < n; i++)
if (f[i] > fleft[i - 1] + f[i])
fleft[i] = f[i];
else
fleft[i] = fleft[i - 1] + f[i];
for (int i = 1; i < n; i++)
fleft[i] = max(fleft[i - 1], fleft[i]);
}

void work2()
{
fright[n - 1] =f[n - 1];
for (int i = n - 2; i >= 0; i--)
if (f[i] > fright[i + 1] + f[i])
fright[i] = f[i];
else
fright[i] = fright[i + 1] + f[i];
for (int i = n - 2; i >= 0; i--)
fright[i] = max(fright[i + 1], fright[i]);
}

void work3()
{
int ans = -1000000000;

for (int i = 0; i < n - 1; i++)
ans = max(ans, fleft[i] + fright[i + 1]);
printf("%d\n", ans);
}

int main()
{
//freopen("t.txt", "r", stdin);
while (scanf("%d", &n), n != 0)
{
input();
work1();
work2();
work3();
}
return 0;
}

转载于:https://www.cnblogs.com/rainydays/archive/2011/05/20/2051996.html