hdu 5115 Dire Wolf（区间dp）

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,  -  feet long and weighing  -  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  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 bi. 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  dire wolves standing in a row, whose basic attacks ai are (, , ), respectively. The extra attacks bi they can provide are (, , ). Thus, the current attacks of them are (, , ). If Matt defeats the second wolf first, he will get  points of damage and the alive wolves’ current attacks become (, ).

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 ( ≤ N ≤ ).

The second line contains N integers ai ( ≤ ai ≤ ), denoting the basic attack of each dire wolf.

The third line contains N integers bi ( ≤ bi ≤ ), 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 ), y is the least damage Matt needs to take.

 

Sample Input

Sample Output

Case #:
Case #: 

Hint

In the first sample, Matt defeats the dire wolves from left to right. He takes  +  +  =  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：Matt遇到了一群狼，他必须解决掉所有的狼才能活，杀掉一只狼 i 时受到的伤害等于a[i]+b[i-1]+b[i+1]，问杀掉所有的狼时受到的最小的伤害。

思路：典型的区间dp

        状态转移方程:当要杀掉狼 k 时，tmp=min(tmp,dp[i][k-1]+dp[k+1][j]+a[k]+b[i-1]+b[j+1]),(i<=k<=j) 

        边界处理，初始化dp[i][i]：for(int i=1;i<=n;i++)  dp[i][i]=a[i]+b[i-1]+b[i+1];

        数组头尾要增加两头"虚狼"，即没有伤害的狼，这样有利于计算的简便。

 

 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<math.h>
 #include<algorithm>
 #include<queue>
 #include<set>
 #include<bitset>
 #include<map>
 #include<vector>
 #include<stdlib.h>
 #include <stack>
 using namespace std;
 #define PI acos(-1.0)
 #define max(a,b) (a) > (b) ? (a) : (b)
 #define min(a,b) (a) < (b) ? (a) : (b)
 #define ll long long
 #define eps 1e-10
 #define MOD 1000000007
 #define N 206
 #define inf 1<<26
 int n;
 int a[N],b[N],dp[N][N];
 int main()
 {
    int t;
    int ac=;
    scanf("%d",&t);
    while(t--){
       scanf("%d",&n);
       a[]=;
       a[n+]=;
       for(int i=;i<=n;i++){
          scanf("%d",&a[i]);
       }
       b[]=;
       b[n+]=;
       for(int i=;i<=n;i++){
          scanf("%d",&b[i]);
       }

       //初始化
       for(int i=;i<=n;i++){
          dp[i][i]=a[i]+b[i-]+b[i+];
       }
       for(int len=;len<=n;len++){
          for(int i=;i+len<=n;i++){
             int j=i+len;
             int tmp=inf;
             for(int k=i;k<=j;k++){
                tmp=min(tmp,dp[i][k-]+dp[k+][j]+a[k]+b[i-]+b[j+]);
             }
             dp[i][j]=tmp;
          }
       }
       printf("Case #%d: ",++ac);
       printf("%d\n",dp[][n]);
    }
     return ;
 }