zoj 3627#類比#枚舉

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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 31 2 3 3 5 42 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

題意 轉自：http://blog.csdn.net/cscj2010/article/details/7819110

題意:n個城市排成一行，每個城市中有vi個金幣。兩個人同時從同一個個城市出發，單位時間能走到相鄰城市。 

•      到達城市擷取金幣不耗時間，且任意時刻兩人距離不可以超過m，問t個時間他們最多能獲得多少金幣。 
• 如果 m >= t * 2,兩個人兩個方向一直走 
• 否則 兩人一直向兩邊走指導相距m，注意，若m為奇數，則某人要停走一天。 
• 然後維持距離同時向左向右枚舉剩餘天數 

 

 

  1 #include<iostream>  2 #include<cstring>  3 #include<cstdlib>  4 #include<cstdio>  5 #include<algorithm>  6 #include<cmath>  7 #include<queue>  8 #include<map>  9 #include<vector> 10  11 #define N 100005 12 #define M 15 13 #define mod 1000000007 14 #define mod2 100000000 15 #define ll long long 16 #define maxi(a,b) (a)>(b)? (a) : (b) 17 #define mini(a,b) (a)<(b)? (a) : (b) 18  19 using namespace std; 20  21 int n,p; 22 ll v[N]; 23 ll tot; 24 ll sum[N]; 25 int m,t; 26 int t1,t2; 27  28 void ini() 29 { 30    int i; 31    tot=0; 32    memset(sum,0,sizeof(sum)); 33    for(i=1;i<=n;i++){ 34         //scanf("%lld",&v[i]); 35         cin>>v[i]; 36         sum[i]=sum[i-1]+v[i]; 37    } 38    scanf("%d%d",&m,&t); 39 } 40  41 void solve() 42 { 43     int i,j,o; 44     if(m/2>=t) 45     { 46         i=max(1,p-t); 47         j=min(n,p+t); 48         tot=sum[j]-sum[i-1]; 49         return ; 50     } 51     t1=min(t,m/2); 52     t2=t-t1; 53     i=max(1,p-t); 54     j=min(n,p+t1); 55     tot=sum[j]-sum[i-1]; 56     for(o=0;o<=t2;o++){ 57         i=max(1,p-t1-o); 58         if(m%2==1 && o!=0){ 59             j=max(p+t1,p+t1+t2-2*o+1); 60             j=min(n,j); 61         } 62  63         else{ 64             j=max(p+t1,p+t1+t2-2*o); 65             j=min(n,j); 66         } 67         tot=max(tot,sum[j]-sum[i-1]); 68     } 69  70     j=min(n,p+t); 71     i=max(1,p-t1); 72     tot=max(tot,sum[j]-sum[i-1]); 73     for(o=0;o<=t2;o++){ 74         if(m%2==1 && o!=0){ 75             i=min(p-t1,p-t1-t2+2*o-1); 76             i=max(1,i); 77         } 78  79         else{ 80             i=min(p-t1,p-t1-t2+2*o); 81             i=max(1,i); 82         } 83  84         j=min(n,p+t1+o); 85        // printf(" o=%d i=%d j=%d sum=%I64d\n",o,i,j,sum[j]-sum[i-1]); 86         tot=max(tot,sum[j]-sum[i-1]); 87     } 88     //tot=v[p]; 89    // i=max(p-t,1); 90     //if(i==1){ 91    //     tot=sum[] 92    // } 93    // j=min(p+1,n); 94  95 } 96  97 int main() 98 { 99     //freopen("data.in","r",stdin);100    // scanf("%d",&T);101    // for(int cnt=1;cnt<=T;cnt++)102    // while(T--)103     while(scanf("%d%d",&n,&p)!=EOF)104     {105         //if(g==0 && b==0 &&s==0) break;106         ini();107         solve();108         //printf("%lld\n",tot);109         cout<<tot<<endl;110     }111 112     return 0;113 }

 

zoj 3627#類比#枚舉