1322: interceptor missile issue (Noip1999)

 

Description [title]

A country in order to defend against enemy missile attacks, the development of a missile interception system, but the interceptor system has a flaw: Although it's the first rounds to reach any height, but each shell and should not be higher than the previous hair height . One day, the radar picked up incoming enemy missiles, because the system is still in beta. Therefore, a system may not intercept all the missiles.

Sequentially input missile flying height (the height of the radar gives a positive integer no greater than 30,000). To calculate the intercept all missiles minimum required number of sets of such a missile interception system is equipped.

[Enter]

n pieces sequentially flying height (1≤n≤1000).

[Output]

To block all equipped with the smallest number of missile systems k.

[Sample input]

389 207 155 300 299 170 158 65

[Sample Output]

2

【prompt】

Input: Missile Height: 432

Output: k = 1 missile interceptors

// Created on 2020/2/12

/*#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <climits>*/
#include <bits/stdc++.h>

using namespace std;

const int idata=1000+5;
int goal[idata];
int step[idata];
int n,m;
int flag;
int cnt;
int ans=1;
int maxx=0;

int main()
{
    int i,j;
    while(scanf("%d",&n)!=EOF)
        goal[cnt++]=n;

    step[1]=goal[0];
    for(i=1;i<cnt;i++)
    {
        flag=0;
        maxx=0;
        for(j=1;j<=ans;j++)
        {
            if(!flag)
            {
                if(step[j]>=goal[i])//不高于前一发，所以“大于等于”
                {
                    flag=j;
                    maxx=step[j];
                }
            }

            if(flag)
            {
                if(maxx>step[j])
                {
                    flag=j;
                    maxx=step[j];
                }
            }
        }
        if(flag)
            step[flag]=goal[i];
        if(!flag)
            step[++ans]=goal[i];
    }

    cout<<ans<<endl;
    /*for(i=1;i<=ans;i++)
    {
        cout<<step[i]<<endl;
    }*/
}