Treasure Hunt II - ZOJ 3627 想法题

Treasure Hunt II

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Time Limit: 2 Seconds       Memory Limit: 65536 KB

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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,"v₁ v₂ ... 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

题意：有n堆金币，两个人一开始在p位置，每个人每天可以移动一个位置，然后两个人的距离不能超过M，问在T天内最多得到的金币是多少。

思路：首先先把两个人张开，能张多大张多大，然后在这种情况下，考虑遍历两边的情况，假设左边走了a步，右边走了b步，那么需要的天数就是2*（a+b）-max（a，b）。然后考虑特殊情况。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
long long num[100010],sum[100010];
int n;
long long solve(int l,int r)
{ if(l<=0)
   l=1;
  if(r>=n)
   r=n;
  return sum[r]-sum[l-1];
}
long long solve2(int l,int r,int t)
{   long long ans;
    int i;
    if(t==0)
    { return solve(l,r);
    }
    if(l<=1)
    { r+=t;
      return solve(1,r);
    }
    if(r>=n)
    { l-=t;
      return solve(l,n);
    }
    else
    { ans=0;
      for(i=0;i<=t/2;i++)
       ans=max(ans,solve(l-i,r+t-i*2));
      for(i=0;i<=t/2;i++)
       ans=max(ans,solve(l-t+i*2,r+i));
      return ans;
    }
}
int main()
{ int i,j,k,l,r,m,t,a,b;
  long long ans;
  sum[0]=0;
  while(~scanf("%d%d",&n,&k))
  { l=k;r=k;
    for(i=1;i<=n;i++)
    { scanf("%lld",&num[i]);
      sum[i]=sum[i-1]+num[i];
    }
    scanf("%d%d",&m,&t);
    if(t<=m/2)
    { l-=t;
      r+=t;
      printf("%lld\n",solve(l,r));
      continue;
    }
    l-=m/2;
    r+=m/2;
    t-=m/2;
    if(t==0)
    { printf("%lld\n",solve(l,r));
      continue;
    }
    if(m%2==0)
     ans=solve2(l,r,t);
    else
     ans=max(solve2(l-1,r,t-1),solve2(l,r+1,t-1));
    printf("%lld\n",ans);
  }
}