# set/map/hash

## Problem
### PA CSES 1629 Movie Festival
https://vjudge.net/problem/CSES-1629
#### Solution

#include <bits/stdc++.h>
#define IOS ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define N 200005
#define int long long
using namespace std;
signed main(){
    IOS
    int n;
    cin >> n;
    pair<int, int> arr[N];
    for(int i = 0;i < n;i++) cin >> arr[i].first >> arr[i].second;
    sort(arr, arr + n, [&](pair<int, int> a, pair<int, int> b){return a.second < b.second;});
    int ans = 1, pre = arr[0].second;
    for(int i = 1;i < n;i++){
        if(arr[i].first >= pre){
            pre = arr[i].second;
            ans++;
        }
    }
    cout << ans << '\n';

    return 0;
}

### PB Sum of Two Values
https://vjudge.net/problem/CSES-1640

#include <bits/stdc++.h>
#define IOS ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;
int n, x;
vector<pair<int, int>> arr;
int main(){
    IOS
    cin >> n >> x;
    for(int i = 0;i < n;i++){
        int num, idx;
        cin >> num;
        arr.push_back({num, i + 1});
    }
    sort(arr.begin(), arr.end());
    // for(int i = 0;i < n;i++) cout << arr[i].first << ' ';
    int ans_x, ans_y;
    bool judge = false;
    for(int l = 0, r = n - 1;l < n;l++){
        while(l < r && arr[l].first + arr[r].first > x) r--;
        if(l != r && arr[l].first + arr[r].first == x){
            judge = true;
            ans_x = arr[l].second;
            ans_y = arr[r].second;
            break;
        }
    }
    if(judge) cout << ans_y << ' ' << ans_x << '\n';
    else cout << "IMPOSSIBLE" << '\n';

    return 0;
}

### PC Kattis Grandpa Bernie
https://vjudge.net/problem/Kattis-grandpabernie

#include <bits/stdc++.h>
#define IOS ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define mod 1000000007
using namespace std;
int main(){
    IOS
    int n;
    unordered_map<string, vector<int> > mp;
    cin >> n;
    vector<string> vec;
    for(int i = 0;i < n;i++){
        string str;
        int tmp;
        cin >> str >> tmp;
        if(mp.count(str)) mp[str].push_back(tmp);
        else{
            vec.push_back(str);
            mp.insert({str, vector<int>()});
            mp[str].push_back(tmp);
        }
    }
    int query;
    cin >> query;
    for(auto i : vec){
        sort(mp[i].begin(), mp[i].end());
    }
    while(query--){
        string str;
        int tmp;
        cin >> str >> tmp;
        cout << mp[str][tmp - 1] << '\n';
    }

    return 0;
}

### PD UVA 12338 Anti-Rhyme Pairs
https://vjudge.net/problem/UVA-12338
待完成
### PE CSES 1076 Sliding Median
https://vjudge.net/problem/CSES-1076
#### Solution

#include <bits/stdc++.h>
#define IOS ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
using namespace std;
typedef long long ll;
void balance(multiset<int, greater<int> > &left, multiset<int, less<int> > &right){
    while(left.size() > right.size()){
        right.insert(*left.begin());
        left.erase(left.begin());
    }
    while(left.size() < right.size()){
        left.insert(*right.begin());
        right.erase(right.begin());
    }
}
int main(){
    IOS
    int n, k, arr[200005];
    cin >> n >> k;
    for(int i = 0;i < n;i++) cin >> arr[i];
    multiset<int, greater<int> > left;
    multiset<int, less<int> > right;
    for(int i = 0;i < n;i++){
        if(!left.size() || arr[i] <= *left.begin()) left.insert(arr[i]);
        else right.insert(arr[i]);
        if(i >= k){
            if(left.size() && arr[i - k] <= *left.begin())
                left.erase(left.find(arr[i - k]));
            else right.erase(right.find(arr[i - k]));
        }
        balance(left, right);
        if(i >= k - 1) cout << *left.begin() << ' ';
    }
    cout << '\n';

    return 0;
}

## Time Complexity Compare
<table>
<thead>
<tr>
<th></th>
<th>set</th>
<th>map</th>
<th>unordered map</th>
</tr>
</thead>
<tbody>
<th>插入</th>
<td></td>
</tbody>
<tbody>
<th>查詢</th>
<td></td>
</tbody>
<tbody>
<th>插入</th>
<td></td>
</tbody>
</table>

## Reference
### PA
給定 n 場電影中的開始時間以及結束時間，要求最多可以看完幾場完整的電影？

pair<int, int> arr[N];
for(int i = 0;i < n;i++) cin >> arr[i].first >> arr[i].second;
sort(arr, arr + n, [&](pair<int, int> a, pair<int, int> b){return a.second < b.second;});
int ans = 1, pre = arr[0].second;
for(int i = 1;i < n;i++){
    if(arr[i].first >= pre){
        pre = arr[i].second;
        ans++;
    }
}

greedy 解，對各場電影的結束時間 sort ，接著用迴圈跑過一次，看各場電影的起頭是否超過上一場電影的結束時間。
## PB
給定 n 個整數求任兩個數字加起來等於 x 。

for(int l = 0, r = n - 1;l < n;l++){
    while(l < r && arr[l].first + arr[r].first > x) r--;
    if(l != r && arr[l].first + arr[r].first == x){
        judge = true;
        ans_x = arr[l].second;
        ans_y = arr[r].second;
        break;
    }
}

雙指針解，藉由雙指針不斷逼近我們要的 x 。