HDU 1024 拦截导弹

 

dp【i】表示为了拦截第i枚导弹最少需要多少个系统

状态转移方程

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <stack>
#include <set>
#include <vector>
//const int maxn = 1e5+5;
#define ll long long
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}

#define MAX INT_MAX
#define FOR(i,a,b) for( int i = a;i <= b;++i)
#define bug cout<<"--------------"<<endl
using namespace std;
int dp[31000],a[31000];
int main()
{
    int n;
    while(cin>>n)
    {

        FOR(i,1,n) cin>>a[i];
        int maxx = 0;
        for(int i=1;i<=n;++i)
        {
            dp[i] = 1;
            for(int j=1;j<i;++j)
            {
                if(a[i] > a[j]) dp[i] = max(dp[j]+1,dp[i]);
            }
            maxx = max(maxx,dp[i]);
        }
        cout<<maxx<<endl;
    }

}

 

来源：https://www.cnblogs.com/jrfr/p/11388935.html