zoj 3627#模拟#枚举

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

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Author: LI, Chao                                                     Contest: ZOJ Monthly, July 2012

题意 转自：http://blog.csdn.net/cscj2010/article/details/7819110

题意:n个城市排成一行，每个城市中有vi个金币。两个人同时从同一个个城市出发，单位时间能走到相邻城市。

• 到达城市获取金币不耗时间，且任意时刻两人距离不可以超过m，问t个时间他们最多能获得多少金币。
• 如果 m >= t * 2,两个人两个方向一直走
• 否则 两人一直向两边走指导相距m，注意，若m为奇数，则某人要停走一天。
• 然后维持距离同时向左向右枚举剩余天数

 #include<iostream>
 #include<cstring>
 #include<cstdlib>
 #include<cstdio>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 #include<map>
 #include<vector>

 #define N 100005
 #define M 15
 #define mod 1000000007
 #define mod2 100000000
 #define ll long long
 #define maxi(a,b) (a)>(b)? (a) : (b)
 #define mini(a,b) (a)<(b)? (a) : (b)

 using namespace std;

 int n,p;
 ll v[N];
 ll tot;
 ll sum[N];
 int m,t;
 int t1,t2;

 void ini()
 {
    int i;
    tot=;
    memset(sum,,sizeof(sum));
    for(i=;i<=n;i++){
         //scanf("%lld",&v[i]);
         cin>>v[i];
         sum[i]=sum[i-]+v[i];
    }
    scanf("%d%d",&m,&t);
 }

 void solve()
 {
     int i,j,o;
     if(m/>=t)
     {
         i=max(,p-t);
         j=min(n,p+t);
         tot=sum[j]-sum[i-];
         return ;
     }
     t1=min(t,m/);
     t2=t-t1;
     i=max(,p-t);
     j=min(n,p+t1);
     tot=sum[j]-sum[i-];
     for(o=;o<=t2;o++){
         i=max(,p-t1-o);
         if(m%== && o!=){
             j=max(p+t1,p+t1+t2-*o+);
             j=min(n,j);
         }

         else{
             j=max(p+t1,p+t1+t2-*o);
             j=min(n,j);
         }
         tot=max(tot,sum[j]-sum[i-]);
     }

     j=min(n,p+t);
     i=max(,p-t1);
     tot=max(tot,sum[j]-sum[i-]);
     for(o=;o<=t2;o++){
         if(m%== && o!=){
             i=min(p-t1,p-t1-t2+*o-);
             i=max(,i);
         }

         else{
             i=min(p-t1,p-t1-t2+*o);
             i=max(,i);
         }

         j=min(n,p+t1+o);
        // printf(" o=%d i=%d j=%d sum=%I64d\n",o,i,j,sum[j]-sum[i-1]);
         tot=max(tot,sum[j]-sum[i-]);
     }
     //tot=v[p];
    // i=max(p-t,1);
     //if(i==1){
    //     tot=sum[]
    // }
    // j=min(p+1,n);

 }

 int main()
 {
     //freopen("data.in","r",stdin);
    // scanf("%d",&T);
    // for(int cnt=1;cnt<=T;cnt++)
    // while(T--)
     while(scanf("%d%d",&n,&p)!=EOF)
     {
         //if(g==0 && b==0 &&s==0) break;
         ini();
         solve();
         //printf("%lld\n",tot);
         cout<<tot<<endl;
     }

     return ;
 }