B. Laurenty and Shop
Time Limit: 1 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/contest/586/problem/B
Description
A little boy Laurenty has been playing his favourite game Nota for quite a while and is now very hungry. The boy wants to make sausage and cheese sandwiches, but first, he needs to buy a sausage and some cheese.
The town where Laurenty lives in is not large. The houses in it are located in two rows, n houses in each row. Laurenty lives in the very last house of the second row. The only shop in town is placed in the first house of the first row.
The first and second rows are separated with the main avenue of the city. The adjacent houses of one row are separated by streets.
Each crosswalk of a street or an avenue has some traffic lights. In order to cross the street, you need to press a button on the traffic light, wait for a while for the green light and cross the street. Different traffic lights can have different waiting time.
The traffic light on the crosswalk from the j-th house of the i-th row to the (j + 1)-th house of the same row has waiting time equal to aij (1 ≤ i ≤ 2, 1 ≤ j ≤ n - 1). For the traffic light on the crossing from the j-th house of one row to the j-th house of another row the waiting time equals bj (1 ≤ j ≤ n). The city doesn't have any other crossings.
The boy wants to get to the store, buy the products and go back. The main avenue of the city is wide enough, so the boy wants to cross it exactly once on the way to the store and exactly once on the way back home. The boy would get bored if he had to walk the same way again, so he wants the way home to be different from the way to the store in at least one crossing.
Figure to the first sample.
Help Laurenty determine the minimum total time he needs to wait at the crossroads.
Input
The first line of the input contains integer n (2 ≤ n ≤ 50) — the number of houses in each row.
Each of the next two lines contains n - 1 space-separated integer — values aij (1 ≤ aij ≤ 100).
The last line contains n space-separated integers bj (1 ≤ bj ≤ 100).
i,Ci，即此题的初始分值、每分钟减少的分值、dxy做这道题需要花费的时间。
Output
Print a single integer — the least total time Laurenty needs to wait at the crossroads, given that he crosses the avenue only once both on his way to the store and on his way back home.
Sample Input
4
1 2 3
3 2 1
3 2 2 3
Sample Output
12
HINT
题意
给你两行房子，你得从第一行第一个走到第二行最后一个，并从第二行最后一个再走回去
每条路都有权值，并且你走过去和走回来的路线要求不一样，问你最小的花费是多少
题解：
首先统计前缀和之后，再暴力枚举中间走的过道是哪些就好了
代码：
#include<stdio.h>
#include<iostream>
#include<math.h>
#include<iostream>
using namespace std; long long a[];
long long b[];
long long c[];
long long suma[];
long long sumb[];
int main()
{
int n;scanf("%d",&n);
for(int i=;i<n;i++)
{
scanf("%d",&a[i]);
suma[i]=a[i]+suma[i-];
}
for(int i=;i<n;i++)
{
scanf("%d",&b[i]);
sumb[i]=b[i]+sumb[i-];
}
for(int i=;i<=n;i++)
scanf("%d",&c[i]);
long long ans = -;
for(int i=;i<=n;i++)
{
for(int j=;j<=n;j++)
{
if(i==j)continue;
if(ans == -)
ans = (suma[i-]+sumb[n-]-sumb[i-]+c[i])+(suma[j-]+sumb[n-]-sumb[j-]+c[j]);
else
ans = min(ans,(suma[i-]+sumb[n-]-sumb[i-]+c[i])+(suma[j-]+sumb[n-]-sumb[j-]+c[j]));
}
}
cout<<ans<<endl;
}