HDU 5115 Dire Wolf (区间DP)

Dire Wolf

                                                                            Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
                                                                                                Total Submission(s): 2761    Accepted Submission(s): 1632

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

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）

题目那么长，我的理解是：有一群狼，第一组数是狼的血量，第二组数是狼可以给相邻的狼增加的血量，你可以杀死其中只狼，他相邻的两只狼都会减去死去的这只狼的对应的第二组数的血量，问杀死全部的狼需要造成最多的伤害。大概就是这个意思，就是第二组数可以加上也可以减去，不死就加上，死了就减去。

听说是模板区间DP，我也是照着区间DP写的。就是枚举区间，用k在i到j间游走，得到状态转移公式: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][j]代表从i到j需要的最小伤害。当杀死第k只狼时，第i-1跟第 j+1就成了相邻的两只狼，血量就会加上对应值。

注意临界值的判断，可以把0位置的狼都设置为0，然后再进行区间DP，这样不用对临界值进行特判。

其他的都是区间DP的知识，也无需多说，只是还是有点不理解区间DP，刷题还是少。

代码实现：

#include <stdio.h>#include <math.h>#include <stdlib.h>#include <string.h>#include <iostream>#include <algorithm>#include <string>#include <queue>#include <stack>#include <vector>#include <set>#include <map>#define mset(a,i) memset(a,i,sizeof(a))using namespace std;typedef long long ll;const int INF=0x3f3f3f3f;const int mod=1e9+7;const int MAX=205;const double eps=1e-6;const double PI=acos(-1);int dir[9][2]={0,1,0,-1,1,0,-1,0, -1,-1,-1,1,1,-1,1,1};inline void SS(char*s){scanf("%s",s);}inline ll SI(){ll x;scanf("%lld",&x);return x;}inline ll P(ll x){printf("%lld\n",x);}inline ll qpow(ll n,ll m){n%=mod;ll ans=1;while(m){if(m%2)ans=(ans*n)%mod;m/=2;n=(n*n)%mod;}return ans;}inline ll inv(ll b){return b==1?1:(mod-mod/b)*inv(mod%b)%mod;}inline ll inv2(ll b){return qpow(b,mod-2);}ll dp[MAX][MAX];int main(){    int n,i,j,k,l,t,flag=1;    int a[MAX],b[MAX];    scanf("%d",&t);    while(t--)    {    a[0]=0;    b[0]=0;        scanf("%d",&n);        for(i=1;i<=n;i++)        scanf("%d",&a[i]);        for(i=1;i<=n;i++)        scanf("%d",&b[i]);        mset(dp,0);        for(l=0;l<=n;l++)        {            for(i=1;i<n-l+1;i++)            {                j=i+l;                dp[i][j]=INF;                for(k=i;k<=j;k++)                {                    dp[i][j]=min(dp[i][j],dp[i][k-1]+dp[k+1][j]+a[k]+b[i-1]+b[j+1]);                }            }        }        printf("Case #%d: %lld\n",flag++,dp[1][n]);    }    return 0;}