BadNeighbors

Problem Statement

The old song declares “Go ahead and hate your neighbor”, and the residents of Onetinville have taken those words to heart. Every resident hates his next-door neighbors on both sides. Nobody is willing to live farther away from the town’s well than his neighbors, so the town has been arranged in a big circle around the well. Unfortunately, the town’s well is in disrepair and needs to be restored. You have been hired to collect donations for the Save Our Well fund.

Each of the town’s residents is willing to donate a certain amount, as specified in the int[] donations, which is listed in clockwise order around the well. However, nobody is willing to contribute to a fund to which his neighbor has also contributed. Next-door neighbors are always listed consecutively in donations, except that the first and last entries in donations are also for next-door neighbors. You must calculate and return the maximum amount of donations that can be collected.

Definition

Class: BadNeighbors
Method: maxDonations
Parameters: int[]
Returns: int
Method signature: int maxDonations(int[] donations)
(be sure your method is public)

Constraints
- donations contains between 2 and 40 elements, inclusive.
- Each element in donations is between 1 and 1000, inclusive.

Examples
0)

{ 10, 3, 2, 5, 7, 8 }

Returns: 19

The maximum donation is 19, achieved by 10+2+7. It would be better to take 10+5+8 except that the 10 and 8 donations are from neighbors.
1)

{ 11, 15 }

Returns: 15

2)

{ 7, 7, 7, 7, 7, 7, 7 }

Returns: 21

3)

{ 1, 2, 3, 4, 5, 1, 2, 3, 4, 5 }

Returns: 16

4)

{ 94, 40, 49, 65, 21, 21, 106, 80, 92, 81, 679, 4, 61,
6, 237, 12, 72, 74, 29, 95, 265, 35, 47, 1, 61, 397,
52, 72, 37, 51, 1, 81, 45, 435, 7, 36, 57, 86, 81, 72 }

Returns: 2926

This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2010, TopCoder, Inc. All rights reserved.
This problem was used for:
2004 TCCC Online Round 4 - Division I, Level One

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题目大意是需要捐款，然后每一个人如果他们的邻居捐款那么他就会不捐款，每个人都有一个对应的捐款值，要求求出最多的捐款数。
我们可以定义f(i)为以i结尾的最大捐款数那么我们可以得到状态转移方程为
f(i)=max[f(i−1),f(i−2)+a(i)] (a(i)表示第i个人的捐款数)
一种不选i即为i−1的值
选了就是没有i−1的值再加上当前的值即f(i−2)+a(i)]

这道题目还有一个关键的地方在于首尾的判断。这里可以把圆环转换成2条线段:
一条包含a(1)而不包含a(n)
另一条不包含a(1)，包含a(n)
对这两条线段进行dp，求出最优值即是答案

#include <bits/stdc++.h>using namespace std;const int maxn=1e5+1;int n,a[maxn],d[maxn],f[maxn],key;char c;int main(){    while(cin>>a[++n]>>c);    for(int i=1;i<n;i++)     f[i]=max(f[i-1],f[i-2]+a[i]);    for(int i=2;i<=n;i++)     d[i]=max(d[i-1],d[i-2]+a[i]);    cout<<max(f[n-1],d[n])<<endl;    return 0;}