HDU-1257

HDU-1257

思路

最少需要的防御系统个数就是最大上升子序列的长度。(Dilworth定理)

这篇blog介绍了Dilworth定理。

Code

$O(n^2)$的LIS

#include <bits/stdc++.h>
using namespace std;
#define _u_u_ ios::sync_with_stdio(false), cin.tie(nullptr)
#define cf int _o_o_;cin>>_o_o_;for (int Case = 1; Case <= _o_o_;Case++)
#define SZ(x) (int)(x.size())
inline void _A_A_();
signed main() {_A_A_();return 0;}

using ll = long long;
// #define int long long
const int mod = 1e9 + 7;
const int maxn = 3e4 + 10;
const int N = 210, M = 5010;
const int inf = 0x3f3f3f3f;

int a[maxn], dp[maxn];

inline void _A_A_() {
    #ifdef LOCAL
    freopen("in.in", "r", stdin);
    #endif
    _u_u_;
    int n;
    while (cin >> n) {
        for (int i = 1;i <= n;i++) cin >> a[i];
        int ans = 1;
        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]);
            }
            ans = max(ans, dp[i]);
        }
        cout << ans << "\n";
    }
}

Code

$O(nlogn)$的LIS，非dp，参考

#include <bits/stdc++.h>
using namespace std;
#define _u_u_ ios::sync_with_stdio(false), cin.tie(nullptr)
#define cf int _o_o_;cin>>_o_o_;for (int Case = 1; Case <= _o_o_;Case++)
#define SZ(x) (int)(x.size())
inline void _A_A_();
signed main() {_A_A_();return 0;}

using ll = long long;
// #define int long long
const int mod = 1e9 + 7;
const int maxn = 3e4 + 10;
const int N = 210, M = 5010;
const int inf = 0x3f3f3f3f;

inline void _A_A_() {
    #ifdef LOCAL
    freopen("in.in", "r", stdin);
    #endif
    _u_u_;
    int n;
    while (cin >> n) {
        vector<int> res;    // 用来存储LIS。
        int a;
        for (int i = 1;i <= n;i++) {
            cin >> a;
            int sz = lower_bound(res.begin(),res.end(), a) - res.begin();
            if (sz == SZ(res)) {
                res.push_back(a);
            }
            else {
                res[sz] = a;
            }
        }
        cout << SZ(res) << "\n";
    }
}