Dire Wolf——HDU5115

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 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 3 dire wolves standing in a row, whose basic attacks ai are (3, 5, 7), respectively. The extra attacks bi 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 ai (0 ≤ ai ≤ 100000), denoting the basic attack of each dire wolf.

The third line contains N integers bi (0 ≤ bi ≤ 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.

题意：

就是有一对狼，每个狼有初始的攻击力，并且还能给左右两边的狼提供攻击力加成，当冒险家杀死一头狼的时候他也会受到这个狼目前攻击力的伤害

实例解析：

3
3 5 7
8 2 0

有三头狼，刚开始第二头狼给他左右两边的狼各加2攻击力，由于第一头狼左边没有狼，所以只给第二头狼加，第三头狼还那样，一系列操作后攻击力为（5，13，9），从左往右杀死狼

1、受到5点攻击，且第二头狼的攻击力加成消失（5，9）

2、受到5点攻击，且第三头狼攻击加成消失（7）

3最后结果5+5+7=17

 

错解：

source就是原有攻击力，add就是增加得攻击力

dp[i][j]=min(dp[i][j],min(dp[i+1][j]+source[i]+add[i+1],dp[i][j-1]+source[j]+add[j-1]));

原本认为就是普通的区间dp，但是在一个父区间中的第一个攻击得人和最后才攻击的人与子区间可能不一样

例如：

3

3  4  5

1 15 3

这样的话最后肯定是先攻击第二个，但是按我们得转移方程那就是从一号和三号中选择一个来先攻击，所以这样就错了<_>

 

正解：

 

dp[i][j]=min(dp[i][j], dp[i][k-1]+dp[k+1][j]+a[k]+b[i-1]+b[j+1])

dp[i][i]=a[i]+b[i-1]+b[j+1];

这个dp[i][j]中的i、j就代表在原给出的序列中杀死第i到j头狼的最小伤害

其中这个k(i<=k<=j)就是枚举那一头狼是最后杀死的

 

这个dp[i][i]就提供了dp[i][j]的结果，所以要确定求dp[i][j]的时候dp[k][k](i<=k<=j)已经求出来了（典型的由部分推整体）

 

代码：

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<algorithm>
 5 #include<vector>
 6 #include<queue>
 7 using namespace std;
 8 typedef long long ll;
 9 const int maxn=205;
10 const int INF=0x3f3f3f3f;
11 const long long inf=0x8080808080808080;
12 const int mod=1000007;
13 int dp[maxn][maxn],v[maxn],w[maxn];
14 int main()
15 {
16     int t,k=0;
17     scanf("%d",&t);
18     while(t--)
19     {
20         k++;
21         int n;
22         scanf("%d",&n);
23         for(int i=1;i<=n;++i)
24         {
25             scanf("%d",&v[i]);
26         }
27         for(int i=1;i<=n;++i)
28         {
29             scanf("%d",&w[i]);
30         }
31         memset(dp,0,sizeof(dp));
32         for(int i=n;i>=1;--i)
33         {
34             for(int j=i;j<=n;++j)
35             {
36                 if(i==j)
37                 {
38                     dp[i][j]=w[i-1]+w[i+1]+v[i];
39                     continue;
40                 }
41                 dp[i][j]=INF;
42                 for(int k=i;k<=j;++k)
43                 {
44                     dp[i][j]=min(dp[i][j],dp[i][k-1]+dp[k+1][j]+w[i-1]+w[j+1]+v[k]);
45                 }
46             }
47         }
48         printf("Case #%d: %d\n",k,dp[1][n]);
49     }
50     return 0;
51 }

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