Dire Wolf

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 1561    Accepted Submission(s):
897

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 ?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.

题目大意(转自别人)：有n头狼排成一排， 每只狼都对相邻的狼的攻击力有加成作用， 每杀死一只狼所受到的伤害为当前狼的攻击力（算上加成的部分）， 被杀死之后的狼对相邻的狼的攻击力的加成会被取消， 同时， 原先与被杀死的狼相邻的两头狼会变成相邻的狼。 要求使得受到的伤害值最小， 求出最小值

为了保证每天至少AC两道题，我无奈又水了一道区间动归题 0.0

 1 #include <cstdio>
 2 #include <cmath>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <queue>
 6 #include <stack>
 7 #include <vector>
 8 #include <iostream>
 9 #include "algorithm"
10 using namespace std;
11 typedef long long LL;
12 const int MAX=205;
13 int t;
14 int n;
15 int a[MAX],b[MAX];
16 int f[MAX][MAX];
17 void init(){
18     int i,j;
19     scanf ("%d",&n);
20     memset(f,0,sizeof(f));
21     memset(a,0,sizeof(a));
22     memset(b,0,sizeof(b));
23     for (i=1;i<=n;i++)
24      scanf("%d",a+i);
25     for (i=1;i<=n;i++)
26      scanf("%d",b+i);
27     for (i=1;i<=n;i++)
28      f[i][i]=a[i]+b[i-1]+b[i+1];
29 }
30 int main(){
31     freopen ("wolf.in","r",stdin);
32     freopen ("wolf.out","w",stdout);
33     int zt(0);
34     scanf("%d",&t);
35     while (t--){
36     init();
37     int len,i,j,k;
38     for (len=2;len<=n;len++)
39      for (i=1;i<=n-len+1;i++)
40      {j=i+len-1;
41       f[i][j]=min(b[i-1]+a[i]+f[i+1][j]+b[j+1],b[i-1]+f[i][j-1]+a[j]+b[j+1]);
42       for (k=i+1;k<j;k++)
43        f[i][j]=min(f[i][j],b[i-1]+f[i][k-1]+a[k]+f[k+1][j]+b[j+1]);
44      }
45     printf("Case #%d: %d\n",++zt,f[1][n]);
46     }
47     return 0;
48 }