文章标题 HDU 5115 ： Dire Wolf （区间DP）

Dire Wolf

Description

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.

题意： 有n头狼排成一列，你想打败他们，每次你只能打败一只，而这一过程会消耗你的血量，消耗的血量为这头狼本生的攻击值和旁边的狼的辅助攻击值，要求的是打败这些狼需要的最少血量。
分析： 一开始看到这个题目，以为是贪心，然后一直得不出最优解，后来才知道是区间ＤＰ（还是太弱鸡了）。定义数组dp[i][j],i，j表示打败第 i 头狼到第 j 头狼之间需要的最少血量，由此可以得出状态转移方程
dp[i][j]=min(dp[i][j],dp[i][k-1]+dp[k+1][j]+a[k]+b[i-1]+b[j+1])
k在【i,j】之间。
代码：

#include<iostream>#include<string>#include<cstdio>#include<cstring>#include<vector>#include<math.h>#include<map>#include<queue> #include<algorithm>using namespace std;const int inf = 0x3f3f3f3f;int dp[205][205];int a[205];int b[205];int main (){    int t,n;    scanf ("%d",&t);    int cnt=1;    while (t--){        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]);        }        a[0]=a[n+1]=b[0]=b[n+1]=0;//第一头没有前驱和最后一头没有后继，就在两边加一个权值为0的前驱和后继。使其符合章台转移方程        for (int i=1;i<=n;i++){//初始化            for (int j=i;j<=n;j++){                dp[i][j]=inf;            }        }        for (int len=0;len<=n;len++){//区间的长度            for (int i=1;i<=n-len;i++){                int j=i+len;                for (int k=i;k<=j;k++){                    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: %d\n",cnt++,dp[1][n]);    }     return 0;}