zoj 3627 F - Treasure Hunt II(贪心)

 

 

 

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu

Submit Status

Description

There are n cities(1, 2, ... ,n) forming a line on the wonderland. city i and city i+1 are adjacent and their distance is 1. Each city has many gold coins. Now, Alice and her friend Bob make a team to go treasure hunting. They starts at city p, and they want to get as many gold coins as possible in T days. Each day Alice and Bob can move to adjacent city or just stay at the place, and their action is independent. While as a team, their max distance can't exceed M.

Input

The input contains multiple cases.
The first line of each case are two integers n, p as above.
The following line contain n interger,"v₁v₂ ... v_n" indicate the gold coins in city i.
The next line is M, T.
(1<=n<=100000, 1<=p<=n, 0<=v_i<=100000, 0<=M<=100000, 0<=T<=100000)

Output

Output the how many gold coins they can collect at most.

Sample Input

6 3
1 2 3 3 5 4
2 1

Sample Output

8

Hint

At day 1: Alice move to city 2, Bob move to city 4.

They can always get the gold coins of the starting city, even if T=0

 

 

贪心题。。一开始先尽量往两边走。。。走到最大距离M...

如果M是奇数的话还要考虑最后一步是谁走（因为最后一步只能一个人走，两个人走就超过M了）

然后分配剩下的时间。。如果往作走了i时间，那么往右就只能走t-2*i(因为还要回来)

然后取所有状态的最大值就好了。。。。

 

具体见代码注释。。。

 

/*************************************************************************
    > File Name: code/zoj/3627.cpp
    > Author: 111qqz
    > Email: rkz2013@126.com 
    > Created Time: 2015年10月24日 星期六 17时00分25秒
 ************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#include<cctype>
                 
#define yn hez111qqz
#define j1 cute111qqz
#define ms(a,x) memset(a,x,sizeof(a))
using namespace std;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
typedef long long LL;
typedef double DB;
const int inf = 0x3f3f3f3f;

const int N=1E5+7;
LL sum[N],v[N];

int n,t,m,p;
LL cal( int l,int r)
{
    l = max(1,l);
    r = min(n,r);//边界处理
    return sum[r]-sum[l-1];
}
int main()
{
  #ifndef  ONLINE_JUDGE 
   freopen("in.txt","r",stdin);
  #endif

   while (scanf("%d %d",&n,&p)!=EOF)
   {
       sum[0] = 0;
       for ( int i = 1 ; i <= n ; i++)
     {
     scanf("%lld",&v[i]);
     sum[i] = sum[i-1] + v[i];
     }
       
       scanf("%d %d",&m,&t);    
       int now = m ;
       int l,r;
       l=r=p;
       while (now>=2 &&t>0)  //在到达距离m之前（m为偶数）或m-1之前（m为奇数）时,同时往两边走
    {
        l--;
        r++;
        t--;
        now = now - 2;

    }
       LL ans = cal(l,r);   //在往两边走的过程中已经用完时间
       LL res;
       int mr,ml;
       for ( int i = 1 ; i <= t ; i++)  //先往右边走，分配时间
    {
        mr = r+i;
        int  tmp = r-m-(t-2*i); //往右走i时间，再回来还需要i时间
                    //此时还剩t-2*i时间，可以让之前再r-m位置的向左走到tmp位置
        ml = min(l,tmp);        //即使tmp比l大，tmp~l的部分再之前往两边走的过程中已经取过了，所以作端点至少为l
                    //换言之，要的是分配i时间走右边的时候能走的最长的那个区间
        res = cal(ml,mr);
        ans = max(ans,res);
       // cout<<"mr:"<<mr<<"tmp:"<<tmp<<" ml:"<<ml<<endl;
    }

       for ( int i = 1 ; i <= t ; i++) //先往左边走，分配时间，同理。
       {
       ml = l-i;
       int tmp = l+m+(t-2*i);
       mr =max(r,tmp);
       res = cal(ml,mr);
       ans = max(ans,res);
       }
       printf("%lld\n",ans);
}
  
   
 #ifndef ONLINE_JUDGE  
  fclose(stdin);
  #endif
    return 0;
}