Dire Wolf（区间DP经典好题）

Link：http://acm.hdu.edu.cn/showproblem.php?pid=5115

Dire Wolf

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 1029    Accepted Submission(s): 582

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.
   
   
   
    
  
  

 

Source

2014ACM/ICPC亚洲区北京站-重现赛（感谢北师和上交）

 

题意：一群狼排成一排，每只狼有一个固定攻击值a[i]，还有附加值他左右与它相邻的狼给它的附加值，因此它对人的攻击总值是固定攻击值+附加值，一个人消灭第i只狼的同时会受到它的总攻击值的伤害，问一个人要按什么顺序消灭全部狼，能使获得受到的攻击值最低。

编程思想：区间DP。

dp[i][j]表示从第i头狼到第j头狼全部被杀死所受到的最小伤害。a[i]表示第i头狼的初始攻击力，b[i]表示第i头狼对相邻狼的加成值。
dp[i][j]=min(dp[i][k-1]+a[k]+dp[k+1][j])+b[i-1]+b[j+1]; (i<k<j&&j>i)，其中k为区间i到j中最后一只杀死的狼。
dp[i][j]=a[i]+b[i-1]+b[j+1];　　(i=j)，其他的详见代码注释。

附：此题与这题 http://blog.youkuaiyun.com/enjoying_science/article/details/49052225 有异曲同工之妙，可参考对比体会，加深对该类DP问题的理解与认识。

模板套路：

dp[i][i]=b[i-1]+a[i]+b[i+1];  i==j,初始化一个点
dp[i][j]=min(b[i-1]+a[i]+b[j+1]+dp[i+1][j],b[j+1]+a[j]+b[i-1]+dp[i][j-1]); 合并一个点和一个区间
dp[i][j]=min(dp[i][j],b[i-1]+a[k]+b[j+1]+dp[i][k-1]+dp[k+1][j]);i+2<=k<=j-2; 合并两个区间

AC code：

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<cstring>
#include<cmath>
#include<vector>
#include<string.h>
#define LL long long
using namespace std;
const int INF=0x3f3f3f3f;
LL dp[222][222];
int a[222],b[222];
/*dp[i][j]表示取出区间i到j的最后一个点的最优值
dp[i][i]=b[i-1]+a[i]+b[i+1];  i==j,初始化一个点
dp[i][j]=min(b[i-1]+a[i]+b[j+1]+dp[i+1][j],b[j+1]+a[j]+b[i-1]+dp[i][j-1]); 合并一个点和一个区间
dp[i][j]=min(dp[i][j],b[i-1]+a[k]+b[j+1]+dp[i][k-1]+dp[k+1][j]);i+2<=k<=j-2; 合并两个区间*/
int main()
{
   //freopen("D:\\in.txt","r",stdin);
   int T,cas,i,j,k,n;
   scanf("%d",&T);
   for(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]);
       }
       b[0]=b[n+1]=0;
       memset(dp,0,sizeof(dp));
       for(i=1;i<=n;i++)
        dp[i][i]=a[i]+b[i-1]+b[i+1];//初始化一个点
       for(i=n-1;i>=1;i--)
       {
           for(j=i+1;j<=n;j++)
           {
              dp[i][j]=min(dp[i+1][j]+b[i-1]+a[i]+b[j+1],dp[i][j-1]+b[i-1]+a[j]+b[j+1]);//合并一个点和一个区间
              for(k=i+1;k<=j-1;k++)
              {
                  dp[i][j]=min(dp[i][j],a[k]+b[i-1]+b[j+1]+dp[i][k-1]+dp[k+1][j]);//合并两个区间
              }
           }
       }
       printf("Case #%d: %I64d\n",cas,dp[1][n]);
   }
    return 0;
}