（HDU 5115）Dire Wolf（区间DP）

Dire Wolf

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 3071    Accepted Submission(s): 1817

Problem 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.

 

区间DP比较简单的一题，要完全消灭标号为1,2,3，，，n的n头狼，用DP的思想来看的话，可以将这个问题分解成更低一级的问题

以DP「i」「j」表示消灭从i到j区间内的狼需要的最小生命值。

那么对于dp【i】【j】，枚举k（i<=k<=j），假设最后一次攻击击杀了标号为k的狼，此时可以推出dp[i][j]=dp[i][k-1]+dp[k+1][j]+a[k]+b[i-1]+b[j+1];

以j-i从小到大的顺序来循环，一直找到dp【1】【n】就是最后的答案。

 

#include<iostream>
#include<cstdio>
#include<vector>
#include<set>
#include<map>
#include<string.h>
#include<cmath>
#include<algorithm>
#include<queue>
#include<stack>
#define LL long long
#define mod 1000000007
#define inf 0x3f3f3f3f

using namespace std;

LL a[210];
LL b[210];
LL dp[210][210];

LL pin(int a,int b)
{
    if(a>b)
        return 0;
    return dp[a][b];
}
int main()
{
    int t;
    scanf("%d",&t);
    for(int p=1;p<=t;p++)
    {
        int n;
        scanf("%d",&n);
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(dp,0,sizeof(dp));

        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(int i=1;i<=n;i++)
            scanf("%d",&b[i]);

        for(int k=0;k<n;k++)
            for(int i=1;i<=n-k;i++)
            {
                if(k==0)
                    dp[i][i]=a[i]+b[i-1]+b[i+1];
                LL ans=inf;
                for(int j=0;j<=k;j++)
                   ans=min(ans,a[i+j]+b[i-1]+b[i+k+1]+pin(i,i+j-1)+pin(i+j+1,i+k));
                dp[i][i+k]=ans;
            }
            printf("Case #%d: %d\n",p,dp[1][n]);
    }
    return 0;
}