Testing the CATCHER
Time Limit: 1000MS   Memory Limit: 30000K
Total Submissions: 13968   Accepted: 5146

Description

A military contractor for the Department of Defense has just completed a series of preliminary tests for a new defensive missile called the CATCHER which is capable of intercepting multiple incoming offensive missiles. The CATCHER is supposed to be a remarkable defensive missile. It can move forward, laterally, and downward at very fast speeds, and it can intercept an offensive missile without being damaged. But it does have one major flaw. Although it can be fired to reach any initial elevation, it has no power to move higher than the last missile that it has intercepted.

The tests which the contractor completed were computer simulations of battlefield and hostile attack conditions. Since they were only preliminary, the simulations tested only the CATCHER's vertical movement capability. In each simulation, the CATCHER was fired at a sequence of offensive missiles which were incoming at fixed time intervals. The only information available to the CATCHER for each incoming missile was its height at the point it could be intercepted and where it appeared in the sequence of missiles. Each incoming missile for a test run is represented in the sequence only once.

The result of each test is reported as the sequence of incoming missiles and the total number of those missiles that are intercepted by the CATCHER in that test.

The General Accounting Office wants to be sure that the simulation test results submitted by the military contractor are attainable, given the constraints of the CATCHER. You must write a program that takes input data representing the pattern of incoming missiles for several different tests and outputs the maximum numbers of missiles that the CATCHER can intercept for those tests. For any incoming missile in a test, the CATCHER is able to intercept it if and only if it satisfies one of these two conditions:

The incoming missile is the first missile to be intercepted in this test.

-or-

The missile was fired after the last missile that was intercepted and it is not higher than the last missile which was intercepted.

Input

The input data for any test consists of a sequence of one or more non-negative integers, all of which are less than or equal to 32,767, representing the heights of the incoming missiles (the test pattern). The last number in each sequence is -1, which signifies the end of data for that particular test and is not considered to represent a missile height. The end of data for the entire input is the number -1 as the first value in a test; it is not considered to be a separate test.

Output

Output for each test consists of a test number (Test #1, Test #2, etc.) and the maximum number of incoming missiles that the CATCHER could possibly intercept for the test. That maximum number appears after an identifying message. There must be at least one blank line between output for successive data sets.

Note: The number of missiles for any given test is not limited. If your solution is based on an inefficient algorithm, it may not execute in the allotted time.

Sample Input

389
207
155
300
299
170
158
65
-1
23
34
21
-1
-1

Sample Output

Test #1:
maximum possible interceptions: 6 Test #2:
maximum possible interceptions: 2

Source

题目有点长,意思也有点扯,答案输入输出也很坑,其实就是求最长递减序列!

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int N=100050;
int a[N],f[N],d[N];
int bsearch(const int *f,int size,const int &a){
int l=0,r=size-1;
while(l<=r){
int mid=(l+r)/2;
if(a<f[mid-1]&&a>=f[mid])return mid;
else if(a>f[mid])r=mid-1;
else l=mid+1;
}
}
int LIS(const int *a,const int &n){
int i,j,size=1;
f[0]=a[0];d[0]=1;
for(i=1;i<n;i++){
if(a[i]>=f[0])j=0;
else if(a[i]<f[size-1])j=size++;
else j=bsearch(f,size,a[i]);
f[j]=a[i];d[i]=j+1;
//for(int k=0;k<size;k++)
// printf(" %d ",f[k]);
//printf("\n");
}
return size;
}
int main()
{
int i,n,t=1;
while(scanf("%d",&a[0])!=EOF&&a[0]!=-1){
for(i=1;;i++)
{
scanf("%d",&a[i]);
if(a[i]==-1)
{
n=i;
break;
}
}
if(t==1)
printf("Test #%d:\n maximum possible interceptions: %d\n",t++,LIS(a,n));
else
printf("\nTest #%d:\n maximum possible interceptions: %d\n",t++,LIS(a,n));
}
return 0;
}

poj1887 Testing the CATCHER的更多相关文章

  1. POJ-1887 Testing the CATCHER(dp,最长下降子序列)

    Testing the CATCHER Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 16515 Accepted: 6082 ...

  2. POJ 1887 Testing the CATCHER

    Testing the CATCHER Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 13396   Accepted: 4 ...

  3. POJ 1887 Testing the CATCHER(LIS的反面 最大递减子序列)

    Language: Default Testing the CATCHER Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 1 ...

  4. POJ 1887:Testing the CATCHER 求递减序列的最大值

    Testing the CATCHER Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 16131   Accepted: 5 ...

  5. Poj 1887 Testing the CATCHER(LIS)

    一.Description A military contractor for the Department of Defense has just completed a series of pre ...

  6. UVa 231 - Testing the CATCHER

    题目大意:一种拦截导弹能拦截多枚导弹,但是它在每次拦截后高度不会再升高,给出导弹的序列,问最多能拦截多少枚导弹? 最长递减子序列问题. #include <cstdio> #include ...

  7. 别人整理的DP大全(转)

    动态规划 动态规划 容易: , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ...

  8. POJ 题目分类(转载)

    Log 2016-3-21 网上找的POJ分类,来源已经不清楚了.百度能百度到一大把.贴一份在博客上,鞭策自己刷题,不能偷懒!! 初期: 一.基本算法: (1)枚举. (poj1753,poj2965 ...

  9. HDU——PKU题目分类

    HDU 模拟题, 枚举1002 1004 1013 1015 1017 1020 1022 1029 1031 1033 1034 1035 1036 1037 1039 1042 1047 1048 ...

随机推荐

  1. ubutun 下webalizer 分析Apache日志

    http://www.webalizer.org/  配置Webalizer 我们可以通过命令行配置Webalizer,也可以通过配置文件进行配置.下面将重点介绍使用配置文件进行配置,该方法使用形式比 ...

  2. 在LINUX的命令提示符及CMD命令提示符中显示时间

    用途之一是可以查看某个命令或程序的执行时间. 一.CMD中显示时间设置 参数说明: $P:当前路径 $G:>(大于号) $T:当前时间,精确到0.01s 实验如下: C:\Users\g4-10 ...

  3. Break the Chocolate(规律)

    Break the Chocolate Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Othe ...

  4. 74HC595的中文资料

    74HC595--具有三态输出锁存功能的8位串行输入.串行/并行输出移位寄存器 本文翻译自NXP的74HC595的datasheet 74HC595和74HCT595是带有存储寄存器和三态输出的8位串 ...

  5. Linux常用命令总结——文件管理

    Linux中的目录 路径:也就是linux中的目录(文件夹)有绝对路径和相对路径 根目录:/ 用户主目录(home directory):位于/home目录下,用户登录时 工作目录(working d ...

  6. 禁用Visual Studio 2013的Browser Link功能

    禁用Visual Studio 2013的Browser Link功能 GET http://localhost:37478/7fd25f8af33f443494e765be19be6240/brow ...

  7. GridView事件分析

    GridView事件分析 (转) P1默认数据绑定过程 编号 事件名称 作用 E1 DataBinding 数据绑定之前触发,在这个事件之前(第一次生成GridView),GridView不存在行数据 ...

  8. java多线程之yield()方法详解

         yiled()方法的作用是放弃当前CPU的资源,将资源让给其它线程,但放弃的时间不确定,有可能刚刚放弃,又马上获得了CPU时间片.下面看一个小例子,看一下具体效果. public stati ...

  9. 工具篇-TraceView

    --- layout: post title: 工具篇-TraceView  description: 让我们远离卡顿和黑屏 2015-10-09 category: blog --- ## 让我们远 ...

  10. Oracle更改数据库文件大小、实时增加文件容量

    --查询数据库文件路径.表空间.大小等 select * from dba_data_files ; --EAST.DBF数据库文件自动扩展20M,可无限扩展 alter database dataf ...