HDU 5115 Dire Wolf （区间DP）2014ICPC 北京站现场赛

Dire Wolf

                                                                          Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
                                                                                                     Total Submission(s): 951    Accepted Submission(s): 536

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.
 

题意：有n只狼，每只狼有基础攻击力ai和buff效果，效果表现为为相邻的两只狼增加bi攻击力。消灭一只狼之后，这只狼相邻的两只狼就变为相邻的狼了，并且原来这只狼的buff消失，相邻狼互相增加buff，求消灭这些狼收到的最小伤害。

分析：看题意，再看N的大小，就能肯定这是一道很裸的区间DP，dp[i][j]代表从消灭i到j的狼所收受的最小伤害，那么dp[i][j]=min{d[i][k-1]+d[k+1][j]+b[i-1]+b[j+1]+a[k]}，这里的a[k]为从i到j的狼的攻击力，因为你要消灭这个区间的狼，这个区间的狼增多的buff就是区间两端相邻的狼的buff，所以要加上b[i-1]+b[j+1]。

#include <cstdio>#include <cstdlib>#include <cstring>#include <cmath>#include <ctime>#include <iostream>#include <algorithm>#include <string>#include <vector>#include <deque>#include <list>#include <set>#include <map>#include <stack>#include <queue>#include <cctype>#include <numeric>#include <iomanip>#include <bitset>#include <sstream>#include <fstream>#define debug "output for debug\n"#define pi (acos(-1.0))#define eps (1e-8)#define inf 0x3f3f3f3f#define ll long long int#define lson l , m , rt << 1#define rson m + 1 , r , rt << 1 | 1using namespace std;const int mod = 1000000007;const int Max = 100005;ll dp[202][202];ll a[202],b[202];ll n,t;int main(){cin>>t;int cnt=1;while(t--)    {        scanf("%lld",&n);        for(int i=1;i<=n;i++)            scanf("%lld",&a[i]);        for(int i=1;i<=n;i++)            scanf("%lld",&b[i]);        memset(dp,0,sizeof dp);        for(int i=1;i<=n;i++)        {            for(int j=i;j<=n;j++)                dp[i][j]=inf;        }        for(int len=0;len<=n;len++)        {            for(int i=1;i<=n-len;i++)            {                int j=i+len;                for(int k=i;k<=j;k++)                {                    dp[i][j]=min(dp[i][k-1]+dp[k+1][j]+a[k]+b[i-1]+b[j+1],dp[i][j]);                }            }        }        printf("Case #%d: %lld\n",cnt++,dp[1][n]);    }    return 0;}

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