B. Laurenty and Shop
Time Limit: 1 Sec ?
Memory Limit: 256 MB
題目連接
http://codeforces.com/contest/586/problem/BDescription
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??,C?i??，即此題的初始分值、每分鐘減少的分值、dxy做這道題需要花費的時間。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
?
題意
給你兩行房子，你得從第一行第一個走到第二行最后一個，并從第二行最后一個再走回去
每條路都有權值，并且你走過去和走回來的路線要求不一樣，問你最小的花費是多少
題解：
首先統計前綴和之后，再暴力枚舉中間走的過道是哪些就好了
代碼：
#include<stdio.h> #include<iostream> #include<math.h> #include<iostream> using namespace std;long long a[501]; long long b[501]; long long c[501]; long long suma[501]; long long sumb[501]; int main() {int n;scanf("%d",&n);for(int i=1;i<n;i++){scanf("%d",&a[i]);suma[i]=a[i]+suma[i-1];}for(int i=1;i<n;i++){scanf("%d",&b[i]);sumb[i]=b[i]+sumb[i-1];}for(int i=1;i<=n;i++)scanf("%d",&c[i]);long long ans = -1;for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){if(i==j)continue;if(ans == -1)ans = (suma[i-1]+sumb[n-1]-sumb[i-1]+c[i])+(suma[j-1]+sumb[n-1]-sumb[j-1]+c[j]);elseans = min(ans,(suma[i-1]+sumb[n-1]-sumb[i-1]+c[i])+(suma[j-1]+sumb[n-1]-sumb[j-1]+c[j]));}}cout<<ans<<endl; }
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