HDU 1257 最少拦截系统

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1257

分析：对于每一个位置（i)，向前找是否存在比它小（或者相等的数）记为j，如果存在，那i必然相较与j得多开一个拦截系统；

#include<iostream>
#include<sstream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<numeric>
#include<cmath>
#include<queue>
#include<vector>
#include<set>
#include<cctype>
#define PI acos(-1.0)
const int INF = 0x3f3f3f3f;
const int NINF = -INF - 1;
const int maxn = 3e4 + 5;
typedef long long ll;
using namespace std;
int a[maxn], dp[maxn];
int main()
{
    int n;
    while (scanf("%d", &n) != EOF)
    {
        int ans = 0;
        for (int i = 0; i < n; ++i)
            scanf("%d", &a[i]);
        for (int i = 0; i < n; ++i) dp[i] = 1;
        for (int i = 0; i < n; ++i)
        {
            for (int j = 0; j < i; ++j)
            {
                if (a[j] <= a[i])
                    dp[i] = max(dp[j] + 1, dp[i]);
            }
            ans = (max(ans, dp[i]));
        }
        //for (int i = 0; i < n; ++i) cout << dp[i] << ' ';
        //cout << endl;
        printf("%d
", ans);
    }
    return 0;
}