题目链接：
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)
#### 问题描述
> 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.
>
#### 输入
> 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.
输出
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.
样例输入
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
样例输出
Case #1: 17
Case #2: 74
题意
n只狼站成一排，每只狼有两个属性：攻击力ai，辅助攻击力bi，一只狼的攻击力为ai+b[i-1]+b[i+1]。你打一只狼会受到ai+b[i-1]+b[i+1]伤害，并且这之狼会死掉，之后相邻的会靠在一起，问你受到最少的伤害打死所有的狼
题解
区间dp
一开始考虑的是[l,r]内，第一只要打死哪只狼，发现死了之后剩下的又会靠在一起。。。
考虑最后一只打死的是哪只狼，就不会有这个问题了，比如你考虑[l,r]内打死的最后一只是i，那么你就发现[l,i],[i,r]完全独立开了，就可以做了。（注意[l,r]中,l和r不能打死）
代码
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf
typedef int LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;
const int INF=0x3f3f3f3f;
const LL INFL=10000000000000000LL;
const double eps=1e-9;
const double PI = acos(-1.0);
//start----------------------------------------------------------------------
const int maxn=222;
int dp[maxn][maxn];
int arr[maxn],brr[maxn];
int n;
int dfs(int l,int r){
if(r-l<=1) return 0;
if(dp[l][r]>=0) return dp[l][r];
int& res=dp[l][r]=INF;
for(int i=l+1;i<r;i++){
res=min(res,dfs(l,i)+dfs(i,r)+brr[l]+brr[r]);
}
return res;
}
int main() {
int tc,kase=0;
scf("%d",&tc);
while(tc--) {
scf("%d",&n);
clr(dp,-1);
for(int i=1;i<=n;i++) scf("%d",&arr[i]);
for(int i=1;i<=n;i++) scf("%d",&brr[i]);
brr[0]=0,brr[n+1]=0;
int ans=dfs(0,n+1);
for(int i=1;i<=n;i++) ans+=arr[i];
prf("Case #%d: %d\n",++kase,ans);
}
return 0;
}
//end-----------------------------------------------------------------------
