The Twin Towers


Time Limit: 2 Seconds      Memory Limit: 65536 KB


Twin towers we see you standing tall, though a building's lost our faith will never fall.

Twin towers the world hears your call, though you're gone it only strengthens our resolve.

We couldn't make it through this without you Lord, through hard times we come together more. ... 



Twin Towers - A Song for America



In memory of the tragic events that unfolded on the morning of September 11, 2001, five-year-old Rosie decids to rebuild a tallest Twin Towers by using the crystals her brother has
collected for years. Will she succeed in building the two towers of the same height?





Input



There are mutiple test cases.

One line forms a test case. The first integer N (N < 100) tells you the number of crystals her brother has collected. Then each of the next N integers describs the height of a certain
crystal.

A negtive N indicats the end.

Note that all crytals are in cube shape. And the total height of crystals is smaller than 2000.

Output



If it is impossible, you would say "Sorry", otherwise tell her the height of the Twin Towers.

Sample Input



4 11 11 11 11

4 1 11 111 1111 

-1

Sample Output



22

Sorry

双塔DP

dp[i][2000] 表示第i个砖块,堆成两个塔的高度差,每个砖块有两个选择要么不用,要么放左边的塔,要么放右边的塔

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h> using namespace std;
int a[105];
int dp[105][4005];
int n;
int main()
{
while(scanf("%d",&n)!=EOF)
{
if(n<0)
break;
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
memset(dp,-1,sizeof(dp));
dp[0][0+2000]=0;
for(int i=1;i<=n;i++)
{
memcpy(dp[i],dp[i-1],sizeof(dp[i]));
for(int j=-1999;j<=1999;j++)
{
if(dp[i-1][j+2000]==-1) continue;
if(j<0)
{
dp[i][j-a[i]+2000]=max(dp[i][j-a[i]+2000],dp[i-1][j+2000]+a[i]);
dp[i][j+a[i]+2000]=max( dp[i][j+a[i]+2000],dp[i-1][j+2000]+max(0,j+a[i]));
}
else
{
dp[i][j+a[i]+2000]=max( dp[i][j+a[i]+2000],dp[i-1][j+2000]+a[i]);
dp[i][j-a[i]+2000]=max(dp[i][j-a[i]+2000],dp[i-1][j+2000]+max(0,a[i]-j));
} }
}
if(dp[n][2000]!=0&&dp[n][2000]!=-1)
printf("%d\n",dp[n][2000]);
else
printf("Sorry\n");
}
return 0;
}
The Twin Towers


Time Limit: 2 Seconds      Memory Limit: 65536 KB


Twin towers we see you standing tall, though a building's lost our faith will never fall.

Twin towers the world hears your call, though you're gone it only strengthens our resolve.

We couldn't make it through this without you Lord, through hard times we come together more. ... 



Twin Towers - A Song for America



In memory of the tragic events that unfolded on the morning of September 11, 2001, five-year-old Rosie decids to rebuild a tallest Twin Towers by using the crystals her brother has
collected for years. Will she succeed in building the two towers of the same height?





Input



There are mutiple test cases.

One line forms a test case. The first integer N (N < 100) tells you the number of crystals her brother has collected. Then each of the next N integers describs the height of a certain
crystal.

A negtive N indicats the end.

Note that all crytals are in cube shape. And the total height of crystals is smaller than 2000.

Output



If it is impossible, you would say "Sorry", otherwise tell her the height of the Twin Towers.

Sample Input



4 11 11 11 11

4 1 11 111 1111 

-1

Sample Output



22

Sorry

ZOJ 2059 The Twin Towers(双塔DP)的更多相关文章

  1. ZOJ 2059 The Twin Towers

    双塔DP. dp[i][j]表示前i个物品,分成两堆(可以不全用),价值之差为j的时候,较小一堆的价值为dp[i][j]. #include<cstdio> #include<cst ...

  2. LightOJ1126 Building Twin Towers(DP)

    题目 Source http://www.lightoj.com/volume_showproblem.php?problem=1126 Description Professor Sofdor Al ...

  3. The Twin Towers zoj2059 DP

    The Twin Towers Time Limit: 2 Seconds      Memory Limit: 65536 KB Twin towers we see you standing ta ...

  4. ZOJ 3331 Process the Tasks 双塔Dp

    用dp[i][j]表示当前安排好了前i个任务,且机器A和机器B完成当前分配到的所有任务的时间差为j(这里j可正可负,实现的时候需要加个offset)时,完成这些任务的最早时间.然后根据j的正负,分别考 ...

  5. lightoj 1126 - Building Twin Towers(dp,递推)

    题目链接:http://www.lightoj.com/volume_showproblem.php?problem=1126 题解:一道基础的dp就是简单的递推可以设dp[height_left][ ...

  6. UVA.10066 The Twin Towers (DP LCS)

    UVA.10066 The Twin Towers (DP LCS) 题意分析 有2座塔,分别由不同长度的石块组成.现在要求移走一些石块,使得这2座塔的高度相同,求高度最大是多少. 问题的实质可以转化 ...

  7. ZOJ 3331 Process the Tasks(双塔DP)

    Process the Tasks Time Limit: 1 Second      Memory Limit: 32768 KB There are two machines A and B. T ...

  8. UVA 10066 The Twin Towers

    裸最长公共子序列 #include<time.h> #include <cstdio> #include <iostream> #include<algori ...

  9. UVA 10066 The Twin Towers(LCS)

    Problem B The Twin Towers Input: standard input Output: standard output Once upon a time, in an anci ...

随机推荐

  1. B9:备忘录模式 Memento

    在不破坏封装性的前提下,捕获一个对象的内部状态,并在该对象之外保存这个状态.这样以后就可以将该对象恢复到原先保存的状态 UML: 示例代码: class Role { private $hp; pri ...

  2. Request.Cookies使用方法分析

    本文章介绍了Request.Cookies的基本的语法和使用方法. 而且通过演示样例分析了Request.Cookies的使用过程. Request.Cookies方法能够检索Cookies 集合中的 ...

  3. js apply和call区别

    <!DOCTYPE html> <html lang="zh"> <head> <meta charset="UTF-8&quo ...

  4. windows 7 提示升级到windows 10补丁

    如果不需要这个提示,可以卸载KB3035583和KB2952664这两个系统更新补丁.   other update:KB2976978   and  KB2977759

  5. scp命令使下用

    先说下背景把,使用ubuntu,没有windows以下到xshell软件管理一下vps.正瞅着须要下载一个chrome(chrominum真的不好用).此刻大谷歌正墙外呢,无奈挂着ss firefox ...

  6. loadrunner两个函数:取参数长度和时间戳函数

    出自中国IT实验室2014-05-23 00:01 1.web_save_param_length 函数 函数原型:int web_save_param_length( const char *Par ...

  7. lucene: 索引建立完后无法查看索引文件中的数据

    索引建立时      1.对原有索引文件进行建立,是可以访问索引文件中的数据的      2.建立新索引文件,必须等建立完毕后,才可以访问,新建立的文件如果没有建立完是不可以被访问的     如果想建 ...

  8. MYSQLMTOP!开源MYSQL监控系统

    原文地址:http://www.lepus.cc/page/opensource

  9. Python3 range()函数

    Python3 range() 函数用法  Python3 内置函数 Python3 range() 函数返回的是一个可迭代对象(类型是对象),而不是列表类型, 所以打印的时候不会打印列表. Pyth ...

  10. Sublime text 3 搭建Python3 IDE

    起因:为了提高编码工作中的体验,Sublime Text:不仅具有华丽的界面,还支持插件扩展机制,用她来写代码,绝对是一种享受. Vim难于上手,Eclipse,VS 体积庞大,即便体积轻巧迅速启动的 ...