F - Treasure Hunt II 模拟题 ，枚举每一种状态

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,"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
***********************************************************************************************************************************************************
题意：两人在有限的距离内和时间内，走一定的格数，求走的格数权值之和最大
***********************************************************************************************************************************************************

 1 #include
 2 #include<string>
 3 #include
 4 #include
 5 #include
 6 #include
 7 #define LL long long
 8 using namespace std;
 9 LL sum[100001];
10 LL val[100001],ans;
11 int n,m,i,j,k,p,t;
12 //该函数由右端决定前段，
13 //两人向左向右走，然后走到相差m时，
14 //再决定同时向左还是向右走，
15 //比较求出最大值，这一过程需要枚举
16 LL find(LL*sum,int p)
17 {
18     LL ans=val[p];
19     int rt,et;
20     for(rt=min(n,p+t);rt>=p;rt--)
21     {
22         int le=max(1,rt-m);
23         le=max(le,p-t);
24         int rest_t=t-max(p-le,rt-p);
25         int l1=max(1,le-rest_t);
26         int r1=min(n,rt+rest_t);
27         ans=max(ans,max(sum[rt]-sum[l1-1],sum[r1]-sum[le-1]));
28     }
29     return ans;
30 }
31 int main()
32 {
33     while(scanf("%d %d",&n,&p)!=EOF)
34     {
35         sum[0]=0;
36         ans=0;
37         for(i=1;i<=n;i++)
38         {
39             scanf("%lld",&val[i]);
40             sum[i]=sum[i-1]+val[i];
41         }
42         scanf("%d %d",&m,&t);
43         LL ans=find(sum,p);
44         for(i=1;i<=n/2;i++)//注意此处一定一定要转换，不然状态不全
45          swap(val[i],val[n-i+1]);
46         sum[0]=0;
47         for(i=1;i<=n;i++)
48          sum[i]=sum[i-1]+val[i];
49         ans=max(ans,find(sum,n-p+1));
50         printf("%lld\n",ans);
51     }
52     return 0;
53 }

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转载于:https://www.cnblogs.com/sdau--codeants/p/3438751.html