hdu 5115 区间dp

链接：戳这里

Dire Wolf

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)

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 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：

n只狼,每只狼有两种属性,a代表狼本身的攻击力,b代表附加攻击力

猎人打死一只狼受到的攻击为：狼本身的攻击力+相邻的狼的附加攻击力

要求猎人杀死所有的狼所受到的攻击最少

思路：

分析了一下看看是否可以贪心做,并不是很好选取先杀的狼

区间dp吧  dp[i][j] 表示杀死区间[i,j]之内所有的狼所受到的攻击最小

枚举区间上的k ,k表示区间间隔点 dp[i][j]=min(dp[i][j],dp[i][k-1]+a[k]+dp[k+1][j]+b[i-1]+b[j+1])

把两个区间合并成一个区间肯定是找到最小的攻击且满足条件的k咯

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<iomanip>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
#define MAX 1000100
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef unsigned long long ull;
#define INF (1ll<<60)-1
using namespace std;
int T,n;
ll dp[1010][1010]; /// dp[i][j] 表示杀死区间[i,j]里面的狼的最小损伤
ll a[1010],b[1010];
int main(){
scanf("%d",&T);
int cas=1;
while(T--){
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%I64d",&a[i]);
for(int i=1;i<=n;i++) scanf("%I64d",&b[i]);
for(int i=1;i<=n;i++){
for(int j=i;j<=n;j++){
dp[i][j]=INF;
}
}
for(int len=1;len<=n;len++){ /// 区间长度
for(int i=1;i<=n-len+1;i++){
int j=i+len-1;
for(int k=i;k<=j;k++){
dp[i][j]=min(dp[i][j],dp[i][k-1]+a[k]+dp[k+1][j]+b[i-1]+b[j+1]);
}
}
}
printf("Case #%d: %I64d\n",cas++,dp[1]
);
}
return 0;
}