Treasure Hunt II
Time 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
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 }