HDU 5115 Dire Wolf（区间DP）

HDU 5115 Dire Wolf（区间DP）。

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_ithey 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 ?rst 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.
        

【题解】

大致题意： 有一排狼，堵在路上，每头狼都有一定的攻击力，而且除了首尾的狼外，你每次攻击任何一头狼，受的伤害处理该狼的攻击力外，还有与该狼相邻的狼的额外攻击力，而首尾的狼的额外攻击力只有来自右边的和左边的狼，现在你要杀掉所有的狼，并使自己受到的伤害值最小，求伤害值。

分析：

典型的区间dp，转移方程：dp[i][j]=min(dp[i][j], dp[i][k-1]+dp[[k+1][j]+a[k]*b[i][*b[j] , 其中a[i]是来自所攻击狼的伤害值，b[i]是该狼为相邻的狼提供的额外攻击力。

注意初始化为击杀每头狼所受的攻击力之和。

【AC代码】

#include
#include
#include
#include
using namespace std;
const int N=205;
const int inf=1e9;
int a[N],b[N];
int dp[N][N];
int m,n;

int main()
{
    int t,cnt=1;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&m);
        for(int i=1;i<=m;++i)
            scanf("%d",&a[i]);
        for(int i=1;i<=m;++i)
            scanf("%d",&b[i]);
        b[0]=b[m+1]=0;//方便计算  不用单独处理

        for(int i=1;i<=m;++i)
            dp[i][i]=a[i]+b[i-1]+b[i+1];

         for(int i=1;i<=m;++i)//遍历从第一只狼杀到第i只狼
        {
            for(int j=1;j<=m-i;++j)
            {
                int k=i+j;
                int s=inf;
                for(int l=j;l<=k;++l)//从第j只狼杀到第k只狼的最少伤害
                {
                    s=min(s,dp[j][l-1]+dp[l+1][k]+a[l]+b[j-1]+b[k+1]);
                }
                dp[j][k]=s;
            }
        }
        printf("Case #%d: %d\n",cnt++,dp[1][m]);
    }
    return 0;
}