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
***********************************************************************************************************************************************************

#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<queue>
#define LL long long
using namespace std;
LL sum[100001];
LL val[100001],ans;
int n,m,i,j,k,p,t;
//该函数由右端决定前段，
//两人向左向右走，然后走到相差m时，
//再决定同时向左还是向右走，
//比较求出最大值，这一过程需要枚举
LL find(LL*sum,int p)
{
    LL ans=val[p];
    int rt,et;
    for(rt=min(n,p+t);rt>=p;rt--)
    {
        int le=max(1,rt-m);
        le=max(le,p-t);
        int rest_t=t-max(p-le,rt-p);
        int l1=max(1,le-rest_t);
        int r1=min(n,rt+rest_t);
        ans=max(ans,max(sum[rt]-sum[l1-1],sum[r1]-sum[le-1]));
    }
    return ans;
}
int main()
{
    while(scanf("%d %d",&n,&p)!=EOF)
    {
        sum[0]=0;
        ans=0;
        for(i=1;i<=n;i++)
        {
            scanf("%lld",&val[i]);
            sum[i]=sum[i-1]+val[i];
        }
        scanf("%d %d",&m,&t);
        LL ans=find(sum,p);
        for(i=1;i<=n/2;i++)//注意此处一定一定要转换，不然状态不全
         swap(val[i],val[n-i+1]);
        sum[0]=0;
        for(i=1;i<=n;i++)
         sum[i]=sum[i-1]+val[i];
        ans=max(ans,find(sum,n-p+1));
        printf("%lld\n",ans);
    }
    return 0;
}