码迷,mamicode.com
首页 > 其他好文 > 详细

hdu 5115 Dire Wolf (区间DP)

时间:2014-12-27 17:32:23      阅读:257      评论:0      收藏:0      [点我收藏+]

标签:

Dire Wolf

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 306    Accepted Submission(s): 179


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 ?rst 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.

令人伤心的一道题。题目意思是有很多狼排成一排,每只狼有一个攻击值a[i]和附加攻击值b[i]。当你消灭一只狼时,你会受到这只狼的攻击值的伤害和它旁边两只狼的附加攻击值的伤害。求消灭所有狼的最小伤害值。

用dp[i][j]表示首先消灭区间[i,j]狼的伤害值,在区间内枚举分割点k属于[i,j-1],k点把区间分成两段,可以先消灭前半段,也可以先消灭后半段,取最优解。

dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]+mint); 

mint=min(b[i-1]-b[k],b[j+1]-b[k+1]);  //先攻击前半段,先攻击后半段

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<vector>
#include<queue>
#include<algorithm>
#include<stdlib.h>
using namespace std;
#define N 205
#define ll __int64
const int inf=0x7fffffff;
int a[N],b[N],dp[N][N];
int main()
{
    int T,i,l,k,cnt=1,n,t;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        a[0]=a[n+1]=b[0]=b[n+1]=0;
        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++)    
            dp[i][i]=a[i]+b[i-1]+b[i+1];
        for(i=2;i<=n;i++)      //枚举区间长度
        {
            for(l=1;l<=n-i+1;l++)  // 每段区间的起点位置
            {
                dp[l][l+i-1]=inf;
                for(k=l;k<l+i-1;k++)   //枚举分割点k
                {
                    t=min(b[l-1]-b[k],b[l+i]-b[k+1]);
                    dp[l][l+i-1]=min(dp[l][l+i-1],dp[l][k]+dp[k+1][l+i-1]+t);
                }
            }
        }
        printf("Case #%d: %d\n",cnt++,dp[1][n]);
    }
    return 0;
}




hdu 5115 Dire Wolf (区间DP)

标签:

原文地址:http://blog.csdn.net/u011721440/article/details/42194759

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!