HDU 5115 区间dp

Dire Wolf

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 2578    Accepted Submission(s): 1509

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

 

   有趣的一道区间dp，看到这种题也想到了是区间dp,一开始想表示dp[i][j]为消灭掉i-j之间的狼所受到的最小伤害，可后来发现一个问题，我们如何表示某个点的伤害能力呢，只要消灭的先后顺序不同受到的伤害就可能会发生改变，如何控制这个变量。我们可以在分割区间时把中间的分割点k看作是这个区间最后一个消灭的狼，[i,j]内k肯定要遍历一遍而且一定有一个是最后杀死的，所以可以这样表示，此时  dp[i][j]=dp[i][k-1]+dp[i][k+1]+a[k]+b[i-1]+b[j+1];

我们把k留在最后就不必在两个子区间内重复计算了，只需要额外加上k造成的伤害，k留在最后所以左边的就是i-1,右边的就是j+1了.

#include<bits/stdc++.h>
using namespace std;
#define inf 0x3f3f3f3f
int a[205],b[205];
int dp[205][205];
int main()
{
    int T,N,m,i,j,k;
    #ifdef ONLINE_JUDGE
    #else
    freopen("in.txt","r",stdin);
    #endif
    cin>>T;
    for(int cas=1;cas<=T;++cas)
    {
        scanf("%d",&N);
        for(i=1;i<=N;++i) scanf("%d",&a[i]);
        for(i=1;i<=N;++i) scanf("%d",&b[i]);
        a[0]=b[0]=a[N+1]=b[N+1]=0;
        memset(dp,0,sizeof(dp));
        dp[0][0]=0;
        for(i=1;i<=N;++i) dp[i][i]=a[i]+b[i-1]+b[i+1];
        for(int len=2;len<=N;++len)
        {
            for(i=1;i+len-1<=N;++i)
            {
                j=i+len-1;
                dp[i][j]=inf;
                for(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",cas,dp[1][N]);
    }
    return 0;
}