POJ3186 Treats for the Cows —— DP
题目链接:http://poj.org/problem?id=3186
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 6548 | Accepted: 3446 |
Description
The treats are interesting for many reasons:
- The treats are numbered 1..N and stored sequentially in single file in a long box that is open at both ends. On any day, FJ can retrieve one treat from either end of his stash of treats.
- Like fine wines and delicious cheeses, the treats improve with age and command greater prices.
- The treats are not uniform: some are better and have higher intrinsic value. Treat i has value v(i) (1 <= v(i) <= 1000).
- Cows pay more for treats that have aged longer: a cow will pay v(i)*a for a treat of age a.
Given the values v(i) of each of the treats lined up in order of the index i in their box, what is the greatest value FJ can receive for them if he orders their sale optimally?
The first treat is sold on day 1 and has age a=1. Each subsequent day increases the age by 1.
Input
Lines 2..N+1: Line i+1 contains the value of treat v(i)
Output
Sample Input
5
1
3
1
5
2
Sample Output
43
Hint
Five treats. On the first day FJ can sell either treat #1 (value 1) or treat #5 (value 2).
FJ sells the treats (values 1, 3, 1, 5, 2) in the following order of indices: 1, 5, 2, 3, 4, making 1x1 + 2x2 + 3x3 + 4x1 + 5x5 = 43.
Source
dp[l][r] = max(dp[l-1][r]+a[l], dp[l][r-1]+a[n+1-r])
当然,还需要注意边界条件:l-1>=0,r-1>=0
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
#define ms(a,b) memset((a),(b),sizeof((a)))
using namespace std;
typedef long long LL;
const double EPS = 1e-;
const int INF = 2e9;
const LL LNF = 2e18;
const int MAXN = 2e3+; int n;
int a[MAXN], dp[MAXN][MAXN]; int main()
{
while(scanf("%d", &n)!=EOF)
{
for(int i = ; i<=n; i++)
scanf("%d", &a[i]); memset(dp, , sizeof(dp));
for(int l = ; l<=n; l++)
for(int r = ; l+r<=n; r++)
{
if(l!=) dp[l][r] = max(dp[l-][r]+(l+r)*a[l], dp[l][r]);
if(r!=) dp[l][r] = max(dp[l][r], dp[l][r-]+(l+r)*a[n+-r]);
} int ans = -INF;
for(int l = ; l<=n; l++)
ans = max(ans, dp[l][n-l]);
printf("%d\n", ans);
}
}
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
#define ms(a,b) memset((a),(b),sizeof((a)))
using namespace std;
typedef long long LL;
const double EPS = 1e-;
const int INF = 2e9;
const LL LNF = 2e18;
const int MAXN = 2e3+; int n;
int a[MAXN], dp[MAXN][MAXN]; int main()
{
while(scanf("%d", &n)!=EOF)
{
for(int i = ; i<=n; i++)
scanf("%d", &a[i]); for(int i = ; i<=n; i++)
dp[i][i] = a[i]*n; for(int len = ; len<=n; len++)
for(int i = ; i+len-<=n; i++)
{
int j = i+len-;
dp[i][j] = max(dp[i+][j]+(n-len+)*a[i], dp[i][j-]+(n-len+)*a[j]);
} printf("%d\n", dp[][n]);
}
}
三.记忆化搜索:
1.上面的两种方法都要考虑枚举顺序的问题,有时比较不好处理。那么可以用记忆化搜索。
2. 思维与方法二一样,只是写法不同。
代码如下:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
#define ms(a,b) memset((a),(b),sizeof((a)))
using namespace std;
typedef long long LL;
const double EPS = 1e-;
const int INF = 2e9;
const LL LNF = 2e18;
const int MAXN = 2e3+; int n;
int a[MAXN], dp[MAXN][MAXN]; int dfs(int l, int r)
{
if(l==r) return n*a[l];
if(dp[l][r]!=-) return dp[l][r];
int k = n-r+l; dp[l][r] = max(k*a[l]+dfs(l+, r), k*a[r]+dfs(l,r-));
return dp[l][r];
} int main()
{
while(scanf("%d", &n)!=EOF)
{
for(int i = ; i<=n; i++)
scanf("%d", &a[i]); memset(dp, -, sizeof(dp));
printf("%d\n", dfs(,n));
}
}
POJ3186 Treats for the Cows —— DP的更多相关文章
- kuangbin专题十二 POJ3186 Treats for the Cows (区间dp)
Treats for the Cows Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 7949 Accepted: 42 ...
- poj3186 Treats for the Cows
http://poj.org/problem?id=3186 Treats for the Cows Time Limit: 1000MS Memory Limit: 65536K Total S ...
- BZOJ 1652: [Usaco2006 Feb]Treats for the Cows( dp )
dp( L , R ) = max( dp( L + 1 , R ) + V_L * ( n - R + L ) , dp( L , R - 1 ) + V_R * ( n - R + L ) ) 边 ...
- poj 3186 Treats for the Cows(dp)
Description FJ has purchased N (1 <= N <= 2000) yummy treats for the cows who get money for gi ...
- poj3186 Treats for the Cows(区间)
题目链接:http://poj.org/problem?id=3186 题意:第一个数是N,接下来N个数,每次只能从队列的首或者尾取出元素. ans=每次取出的值*出列的序号.求ans的最大值. 样例 ...
- POJ3186:Treats for the Cows(区间DP)
Description FJ has purchased N (1 <= N <= 2000) yummy treats for the cows who get money for gi ...
- 【POJ - 3186】Treats for the Cows (区间dp)
Treats for the Cows 先搬中文 Descriptions: 给你n个数字v(1),v(2),...,v(n-1),v(n),每次你可以取出最左端的数字或者取出最右端的数字,一共取n次 ...
- poj 3186 Treats for the Cows(区间dp)
Description FJ has purchased N (1 <= N <= 2000) yummy treats for the cows who get money for gi ...
- (区间dp + 记忆化搜索)Treats for the Cows (POJ 3186)
http://poj.org/problem?id=3186 Description FJ has purchased N (1 <= N <= 2000) yummy treats ...
随机推荐
- Java学习之并发多线程理解
1.线程简介: 世间万物会同时完成很多工作,如人体同时进行呼吸.血液循环.思考问题等活动,用户既可以使用计算机听歌也可以使用它打印文件,而这些活动完全可以同时进行,这种思想在Java中称为并发,而将并 ...
- Java学习之集合框架的迭代器--Iteratorjk及ListItertor接口
通常情况下,你会希望遍历一个集合中的元素.例如,显示集合中的每个元素.一般遍历数组都是采用for循环或者增强for,这两个方法也可以用在集合框架,但是还有一种方法是采用迭代器遍历集合框架,它是一个对象 ...
- Fidder详解-工具简介(保存会话、decode解码、Repaly、自定义会话框、隐藏会话、会话排序)
前言 本文会对Fidder这款工具的一些重要功能,进行详细讲解,带大家进入Fidder的世界,本文会让你明白,Fidder不仅是一个抓包分析工具,也是一个请求发送工具,更加可以当作为Mock Serv ...
- 【java 理论篇 2】J2EE的13种规范
导读:看完了J2EE的视频,没有什么技术实践,现在就从理论上说明一下J2EE的13种规范,以及现在的自己对它的一个理解.可能会有偏差,但是,算是做为目前的一个记录. 一.13种规范 1.1.JDBC( ...
- 银河英雄传说(codevs 1540)
题目描述 Description 公元五八○一年,地球居民迁移至金牛座α第二行星,在那里发表银河联邦创立宣言,同年改元为宇宙历元年,并开始向银河系深处拓展. 宇宙历七九九年,银河系的两大军事集团在巴米 ...
- linux下部署一个JavaEE项目的简单步骤
部署项目的前提是准备好环境,包括:Java运行环境(JDK),Tomcat,Mysql数据库 1.首先将东西上传到服务器,我是在windows安装ssh工具(下载安装git即可http://gitfo ...
- 1597: [Usaco2008 Mar]土地购买 [ dp+斜率优化 ] 未完
传送门 1597: [Usaco2008 Mar]土地购买 Time Limit: 10 Sec Memory Limit: 162 MBSubmit: 1979 Solved: 705[Subm ...
- iOS 调用系统相册 相机 时,显示中文标题
解决手机语言已经设置显示中文 在调用系统相册.相机界面 时显示英文问题, 在 info.plist里面添加Localized resources can be mixed YES 表示是否允许应用程序 ...
- Python从文件中读取字符串,用正则表达式匹配中文字符的问题
2013-07-27 21:01:37| 在Windows下,用Python从.txt文件中读取字符串,并用正则表达式匹配中文,在网上看了方法,用的时候发现中文没有被匹配. ...
- Java模拟斗地主(实现大小排序)
import java.util.Arrays; import java.util.Collections; import java.util.HashMap; import java.util.Li ...