(简单dp 水过) poj 1887 Testing the CATCHER

Testing the CATCHER

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 8793		Accepted: 3139

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

38920715530029917015865-1233421-1-1

Sample Output

Test #1:  

maximum possible interceptions: 6

Test #2:  

maximum possible interceptions: 2

/*
题目大意：求最长下降序列。
题目解答：
		  1：最优子结构　　　　　
		：设序列X={x1,x2,x3,...xm}的最大下降子序列为Y={y1,y2,y3,...yn};
　　　　：若yn==xk,则子序列X‘’={x1,x2,x3,x4...xk-1};
　　　　：证明：如果x‘’不是最优解(即不是前k-1子序列的最长下降子序列),那么必定存在一个x“
		：序列比x''更优,那么可以导出X的解并不是最优解，那应该是x'''+1。
		：结果跟假设条件矛盾。说明这个问题有最优子结构.　　　　
		  2：子问题的递归结构　　　　　　
		：X序列={x1,x2,x3...xi}；当i=1的时候,opt[i]=1;其他情况下，有最优子结构性质可建立递归关系　　　　　　
		：opt[i]=max(opt[j]+1),num[i] < num[j] && 0<=j<i<m ps：这个算法的复杂度o(n^2),也是效率比较差的，
		： 还有一种效率高的情见poj 1631 Bridging signals 复杂度为o(nlogn);ps：注意格式，要每一个case输出一个空行.

Source Code

Problem: 1887  User: wawadimu 
Memory: 204K  Time: 63MS 
Language: C++  Result: Accepted 

Source Code
*/ 
#include<iostream>
#include<queue>
using namespace std;

#define maxn 32768
int num[maxn];
int opt[maxn];
int main()
{
	//freopen("1887.txt","r",stdin);
	int i,j,k,tmp;
	int cnt,Case=0;;
	while(scanf("%d",&tmp)!=EOF && tmp!=-1)
	{
		Case++;
		cnt=0;
		num[cnt++]=tmp;
		while(scanf("%d",&tmp)!=EOF && tmp!=-1)
		{
			num[cnt++]=tmp;
		}
		int max=1;
		for(i=0;i<cnt;i++)
		{
			opt[i]=1;
			for(j=0;j<i;j++)
			{
				if(num[i] < num[j] && opt[i] <= opt[j])
				{
					opt[i]=opt[j]+1;
				}
			}
			if(max < opt[i]) max=opt[i];
		}
		printf("Test #%d:/n  maximum possible interceptions: %d/n/n",Case,max);
	}
	return 1;
}