zoj 3627#模拟#枚举

Treasure Hunt IITime Limit: 2 Seconds Memory Limit: 65536 KB
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 ... vn" indicate the gold coins in city i. The next line is M, T. (1<=n<=100000, 1<=p<=n, 0<=vi<=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

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
;
ll tot;
ll sum
;
int m,t;
int t1,t2;

void ini()
{
int i;
tot=0;
memset(sum,0,sizeof(sum));
for(i=1;i<=n;i++){
//scanf("%lld",&v[i]);
cin>>v[i];
sum[i]=sum[i-1]+v[i];
}
scanf("%d%d",&m,&t);
}

void solve()
{
int i,j,o;
if(m/2>=t)
{
i=max(1,p-t);
j=min(n,p+t);
tot=sum[j]-sum[i-1];
return ;
}
t1=min(t,m/2);
t2=t-t1;
i=max(1,p-t);
j=min(n,p+t1);
tot=sum[j]-sum[i-1];
for(o=0;o<=t2;o++){
i=max(1,p-t1-o);
if(m%2==1 && o!=0){
j=max(p+t1,p+t1+t2-2*o+1);
j=min(n,j);
}

else{
j=max(p+t1,p+t1+t2-2*o);
j=min(n,j);
}
tot=max(tot,sum[j]-sum[i-1]);
}

j=min(n,p+t);
i=max(1,p-t1);
tot=max(tot,sum[j]-sum[i-1]);
for(o=0;o<=t2;o++){
if(m%2==1 && o!=0){
i=min(p-t1,p-t1-t2+2*o-1);
i=max(1,i);
}

else{
i=min(p-t1,p-t1-t2+2*o);
i=max(1,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-1]);
}
//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 0;
}