HDU5115 Dire Wolf（区间DP）

Dire Wolf

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

Total Submission(s): 1257    Accepted Submission(s): 712

Problem Description3

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

233 5 78 2 0101 3 5 7 9 2 4 6 8 109 4 1 2 1 2 1 4 5 1

 

Sample Output

Case #1: 17Case #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.
 

 

Source

2014ACM/ICPC亚洲区北京站-重现赛（感谢北师和上交）  

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5115

题意：

有n只狼，每只狼有两种属性，一种攻击力（第一行）一种附加值（第二行），我们没杀一只狼，那么我们受到的伤害值为

这只狼的攻击值与它旁边的两只狼的附加值的和，求把所有狼都杀光受到的最小的伤害值。

区间DP的定义：

区间动态规划问题一般都是考虑，对于每段区间，他们的最优值都是由几段更小区间的最优值得到，是分治思想的一种应用，将一个区间问题不断划分为更小的区间直至一个元素组成的区间，枚举他们的组合 ，求合并后的最优值。

设F[i,j]（1<=i<=j<=n）表示区间[i,j]内的数字相加的最小代价
最小区间F[i,i]=0（一个数字无法合并，∴代价为0）

每次用变量k（i<=k<=j-1）将区间分为[i,k]和[k+1,j]两段

For p:=1 to n do // p是区间长度，作为阶段。 
for i:=1 to n do // i是穷举的区间的起点
begin
j:=i+p-1; // j是 区间的终点，这样所有的区间就穷举完毕
if j>n then break; // 这个if很关键。
for k:= i to j-1 do // 状态转移，去推出 f[i,j]
f[i , j]= max{f[ i,k]+ f[k+1,j]+ w[i,j] } 
end; 
这个结构必须记好，这是区间动态规划的代码结构。

AC代码：

#include <stdio.h>#include <string.h>int main(){    int t,n,i,j,length;    int start,mid,end;    int a[210],b[210];    int dp[210][210];    int times=0;    scanf("%d",&t);    while(t--)    {        memset(dp,0,sizeof(dp));        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));        scanf("%d",&n);        for(i=1; i<=n; i++)        {            scanf("%d",&a[i]);        }        for(i=1; i<=n; i++)        {            scanf("%d",&b[i]);        }        for (i=1; i<=n; i++)        {            for (j=i; j<=n; j++)                dp[i][j]=99999999;        }        for (length=0;length<=n;length++)/**区间长度*/        {            for (start=1; start<n+1-length; start++)/**区间的起点*/            {                end=start+length;/**终点(起点+长度==终点)*/                for (mid=start; mid<=end; mid++)/**中间点*/                {                    if (dp[start][mid-1]+dp[mid+1][end]+a[mid]+b[start-1]+b[end+1]<dp[start][end])                    {                        dp[start][end]=dp[start][mid-1]+dp[mid+1][end]+a[mid]+b[start-1]+b[end+1];                    }                }            }        }        printf("Case #%d: %ld\n",++times,dp[1][n]);    }    return 0;}