zoukankan      html  css  js  c++  java
  • HDU5115 Dire Wolf

    Time Limit: 5000MS   Memory Limit: 512000KB   64bit IO Format: %I64d & %I64u

    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

    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.

    Source

     
    迷之区间DP。
    因为每次攻击都要考虑来自目标左右两侧的伤害,所以区间长度为2的时候有些难处理。
    只要在初始化的时候特殊处理dp数组即可。
    具体看代码:
     1 /*by SilverN*/
     2 #include<iostream>
     3 #include<algorithm>
     4 #include<cstring>
     5 #include<cstdio>
     6 #include<cmath>
     7 using namespace std;
     8 const int mxn=240;
     9 int read(){
    10     int x=0,f=1;char ch=getchar();
    11     while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    12     while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();}
    13     return x*f;
    14 }
    15 int T;
    16 int n;
    17 int a[mxn],b[mxn];
    18 int f[mxn][mxn];
    19 int main(){
    20     T=read();
    21     int i,j;
    22     int cas=0;
    23     while(T--){
    24 //        memset(f,0x3f,sizeof f);
    25         memset(f,0,sizeof f);
    26         n=read();
    27         for(i=1;i<=n;i++){a[i]=read();}
    28         for(i=1;i<=n;i++){b[i]=read();}
    29         a[n+1]=b[n+1]=0;
    30         for(i=1;i<=n;i++)
    31             for(j=i;j<=n;j++)
    32                 if(i==j)f[i][i]=a[i]+b[i-1]+b[i+1];
    33                 else f[i][j]=0x3f3f3f3f;
    34         for(int st=1;st<=n;st++)
    35           for(i=1;i<=n-st;i++){
    36               j=i+st;
    37               for(int k=i;k<=j;k++){
    38                   f[i][j]=min(f[i][j],f[i][k-1]+f[k+1][j]+a[k]+b[i-1]+b[j+1]);
    39               }
    40 //            printf("%d %d %d
    ",i,j,f[i][j]);
    41           }
    42         
    43         printf("Case #%d: %d
    ",++cas,f[1][n]);
    44     }
    45     return 0;
    46 }
  • 相关阅读:
    基于antd封装一个高可用form组件 减少cv代码导致的bug
    写一个umi插件 自动生成代码 解放cv的双手
    jquery.validate的一个bug
    ef6+mysql的bug
    angularJS使用ocLazyLoad实现js延迟加载
    postgresql,postgis,geoserver 发布地图服务,并用.net mvc openlayers3进行显示
    EF框架操作postgresql,实现WKT类型坐标的插入,查询,以及判断是否相交
    jquery validate 之多tab页同时校验问题
    js去除空格12种方法
    个人工具清单
  • 原文地址:https://www.cnblogs.com/SilverNebula/p/6022568.html
Copyright © 2011-2022 走看看