(DP) codeforces 358D

D. Dima and Hares

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Dima liked the present he got from Inna very much. He liked the present he got from Seryozha even more.

Dima felt so grateful to Inna about the present that he decided to buy her n hares. Inna was very happy. She lined up the hares in a row, numbered them from 1 to n from left to right and started feeding them with carrots. Inna was determined to feed each hare exactly once. But in what order should she feed them?

Inna noticed that each hare radiates joy when she feeds it. And the joy of the specific hare depends on whether Inna fed its adjacent hares before feeding it. Inna knows how much joy a hare radiates if it eats when either both of his adjacent hares are hungry, or one of the adjacent hares is full (that is, has been fed), or both of the adjacent hares are full. Please note that hares number 1 and n don't have a left and a right-adjacent hare correspondingly, so they can never have two full adjacent hares.

Help Inna maximize the total joy the hares radiate. :)

Input

The first line of the input contains integer n (1 ≤ n ≤ 3000) — the number of hares. Then three lines follow, each line has n integers. The first line contains integers a1 a2 ... an. The second line contains b1, b2, ..., bn. The third line contains c1, c2, ..., cn. The following limits are fulfilled: 0 ≤ ai, bi, ci ≤ 105.

Number ai in the first line shows the joy that hare number i gets if his adjacent hares are both hungry. Number bi in the second line shows the joy that hare number i radiates if he has exactly one full adjacent hare. Number сi in the third line shows the joy that hare number i radiates if both his adjacent hares are full.

Output

In a single line, print the maximum possible total joy of the hares Inna can get by feeding them.

Sample test(s)

input

4
1 2 3 4
4 3 2 1
0 1 1 0

output

13

input

7
8 5 7 6 1 8 9
2 7 9 5 4 3 1
2 3 3 4 1 1 3

output

44

input

3
1 1 1
1 2 1
1 1 1

output

4

题目大意：从1到n的位置取数，取数的得到值与周围的数有没有取过有关，所有数都要取，求最终得到的最大结果

解题思路：dp题，转移方程如下

dp[i][0]=max(dp[i-1][0]+b[i-1],dp[i-1][1]+c[i-1])

dp[i][1]=max(dp[i-1][0]+a[i-1],dp[i-1][1]+b[i-1])

a,b,c分别表示周围没有数，有一个数，有两个数取过的情况。

dp[i][0]表示取i个位置时，i-1没取过的情况。（实际取数的情况，先i，再i-1）

dp[i][1]表示取i个位置时，i-1取过的情况。（实际取数的情况，先i-1，再i）

囧。。。窝就是个DP渣渣

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<cstdlib>
using namespace std;
int a[3005],b[3005],c[3005],dp[3005][2],n;
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    for(int i=1;i<=n;i++)
        scanf("%d",&b[i]);
    for(int i=1;i<=n;i++)
        scanf("%d",&c[i]);
    if(n==1)
    {
        printf("%d
",a[1]);
        return 0;
    }
    dp[2][0]=b[1];
    dp[2][1]=a[1];
    for(int i=3;i<=n;i++)
    {
        dp[i][0]=max(dp[i-1][0]+b[i-1],dp[i-1][1]+c[i-1]);
        dp[i][1]=max(dp[i-1][0]+a[i-1],dp[i-1][1]+b[i-1]);
    }
    int ans=max(dp[n][0]+a[n],dp[n][1]+b[n]);
    printf("%d
",ans);
    return 0;
}