Dire wolves, also known as Dark wolves, are extraordinarily large and powerful wolves. Many, if not all, Dire Wolves appear to originate from Draenor.
Dire wolves look like normal wolves, but these creatures are of nearly twice the size. These powerful beasts, 8 - 9 feet long and weighing 600 - 800 pounds, are the most well-known orc mounts. As tall as a man, these great wolves have long tusked jaws that look like they could snap an iron bar. They have burning red eyes. Dire wolves are mottled gray or black in color. Dire wolves thrive in the northern regions of Kalimdor and in Mulgore.
Dire wolves are efficient pack hunters that kill anything they catch. They prefer to attack in packs, surrounding and flanking a foe when they can.
— Wowpedia, Your wiki guide to the World of Warcra
Matt, an adventurer from the Eastern Kingdoms, meets a pack of dire wolves. There are N wolves standing in a row (numbered with 1 to N from left to right). Matt has to defeat all of them to survive.
Once Matt defeats a dire wolf, he will take some damage which is equal to the wolf’s current attack. As gregarious beasts, each dire wolf i can increase its adjacent wolves’ attack by b i. Thus, each dire wolf i’s current attack consists of two parts, its basic attack ai and the extra attack provided by the current adjacent wolves. The increase of attack is temporary. Once a wolf is defeated, its adjacent wolves will no longer get extra attack from it. However, these two wolves (if exist) will become adjacent to each other now.
For example, suppose there are 3 dire wolves standing in a row, whose basic attacks ai are (3, 5, 7), respectively. The extra attacks b i they can provide are (8, 2, 0). Thus, the current attacks of them are (5, 13, 9). If Matt defeats the second wolf first, he will get 13 points of damage and the alive wolves’ current attacks become (3, 15).
As an alert and resourceful adventurer, Matt can decide the order of the dire wolves he defeats. Therefore, he wants to know the least damage he has to take to defeat all the wolves.
Input
The first line contains only one integer T , which indicates the number of test cases. For each test case, the first line contains only one integer N (2 ≤ N ≤ 200).
The second line contains N integers a i (0 ≤ a i ≤ 100000), denoting the basic attack of each dire wolf.
The third line contains N integers b i (0 ≤ b i ≤ 50000), denoting the extra attack each dire wolf can provide.
Output
For each test case, output a single line “Case #x: y”, where x is the case number (starting from 1), y is the least damage Matt needs to take.
Sample Input
2
3
3 5 7
8 2 0
10
1 3 5 7 9 2 4 6 8 10
9 4 1 2 1 2 1 4 5 1
Sample Output
Case #1: 17
Case #2: 74
Hint
In the first sample, Matt defeats the dire wolves from left to right. He takes 5 + 5 + 7 = 17 points of damage which is the least damage he has to take.
題目大意：
有一排狼，每隻狼都有自己的基礎攻擊值和額外攻擊值，當你去殺一頭狼的時候，你會受到狼本身的基礎攻擊值，和它左右兩邊狼的額外攻擊值，問你，當你把所有狼都殺完的時候，最小受到的攻擊值是多少
解題思路：
根據題目給的樣例會發現，只要求怎麼使額外攻擊值最小就行，其他攻擊值是一定的，而且，先殺額外攻擊值最大的，然後是殺兩側的，會使得額外攻擊值最小
做法是區間DP:
dp[i][j]代表殺光區間i-j所有的狼，所受到的最小攻擊值,k是區間i-j裏面最後殺的那隻狼
轉移方程：
dp[i][j]=min(dp[i][j],dp[i][k-1]+dp[k+1][j]+a[k]+b[i-1]+b[j+1])///a[k]是基礎攻擊，b[i-1]是i-j區間的左邊的值，b[j+1]是i-j區間的的右邊的值，因爲k是i-j區間最後殺的，也就是說i-j區間內只有它了，所以它左右緊湊過來的，應該是開始的時候區間i-j兩邊的值，也就是i-1和j+1
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
using namespace std;
const int INF=0x3f3f3f3f;
int dp[210][210];///dp[i][j]代表全部殺光第i只狼到第j只狼所受到的最小攻擊
int a[210],b[210];///a是基本攻擊，b是額外攻擊
int main()
{
int t,cas,n;
scanf("%d",&t);
for(int cas=1;cas<=t;cas++)
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
for(int j=1;j<=n;j++)
scanf("%d",&b[j]);
a[0]=b[0]=a[n+1]=b[n+1]=0;///初始化邊界情況
memset(dp,INF,sizeof(dp));
for(int i=1;i<=n;i++)
dp[i][i]=a[i]+b[i-1]+b[i+1];///區間長度爲1的時候的情況
for(int len=2;len<=n;len++)///枚舉所有可能的區間長度
{
for(int i=1;i<=n-len+1;i++)///枚舉所有可能的起點
{
int j=i+len-1;///j代表區間的右端點///即區間[i,j]
dp[i][j]=min(dp[i+1][j]+a[i]+b[i-1]+b[j+1],dp[i][j-1]+a[j]+b[i-1]+b[j+1]);///先考慮左右端點時候的特殊情況
for(int k=i+1;k<j;k++)///k是最後一隻殺的狼
dp[i][j]=min(dp[i][j],dp[i][k-1]+dp[k+1][j]+a[k]+b[i-1]+b[j+1]);
}
}
printf("Case #%d: ",cas);
printf("%d\n",dp[1][n]);
}
}
