杭电杯第三场

个人题解，欢迎指正

题目变得好难，我也一直在推A题，还没推出来，最后一个多小时来看P1009，还好写出来了

P1003

思路：

签到题

Code：

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<int,int> pii;

#define IOS ios::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr)
#define lowbit(x)   ((x)&(-x))
#define fi first
#define se second
#define pb push_back
#define cf(_) int _;cin >> _;while(_--)

template <typename T> bool chkMax(T &x, T y) { return (y > x) ? x = y, 1 : 0; }
template <typename T> bool chkMin(T &x, T y) { return (y < x) ? x = y, 1 : 0; }

template <typename T> void inline read(T &x) {
    int f = 1; x = 0; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') f = -1; c = getchar(); }
    while (c <= '9' && c >= '0') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    x *= f;
}

int main()
{
    
    int T;
    cin >> T;
    getchar();
    while(T--)
    {
        
        string s;

        getline(cin,s);
        string ans;
        bool flag = false;
        // cout << s << endl;
        for(int i = 0;i < s.size();i++)
        {
            if(!i) ans.pb(s[i]);
            else
            {
                // cout << s[i] ;
                if(s[i] == ' ') {
                    flag = true;
                    continue;
                }
                if(flag)
                {
                    // cout << s[i];
                    ans.pb(s[i]);
                    flag = false;
                }
            } 
        }
        for(int i = 0;i < ans.size();i++)
        {
            cout << char(ans[i] - 32);
        }
        cout << endl;
    }
    return 0;
}

P1009

思路：

将$l,r$排序后开始遍历，让开始的$R$为第一个$r$,然后将$l$小于$R$的数都加入一个优先队列中，同时将$R$更新为当前队列中最小的$r$,然后将数据弹出，要么队列为空，要么剩余带的快递数量为$0$。

Code：

#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define WA return 0;
#define ptn putchar('\n')
using namespace std;

inline ll read() { ll x = 0, z = 1; char c = getchar(); while (!isdigit(c)) { if (c == '-')z = -1; c = getchar(); }while (isdigit(c)) { x = (x << 1) + (x << 3) + (c ^ 48); c = getchar(); }return z * x; }
inline void writ(ll x) { if (x < 0) { putchar('-'); x = (~x) + 1; }if (x > 9)writ(x / 10); putchar(x - x / 10 * 10 + 48); }

#define pii pair<int,int>
const ll INF = 1e18+7;


void Qingtuan() {
    int n = read(), k = read();
    priority_queue<pii,vector<pii>,greater<pii>>pq;			//改为数组排序也可
    pii t;
    for (int i = 1; i <= n; i++)
    {
        t.first = read(); t.second = read();
        pq.push(t);
    }
    priority_queue<int,vector<int>,greater<int>>p;
    int res=0;
    int r;
    int m;
    while (!p.empty()||!pq.empty())
    {
        m=k;
        if(p.empty()){
            t=pq.top();
            r=t.second;
        }else{
            r=p.top();
        }
        while (!pq.empty())
        {
            t=pq.top();
            if(t.first<=r){
                pq.pop();p.push(t.second);
                r=min(r,t.second);
            }else{
                break;
            }
        }
        
        while (!p.empty()&&m)
        {
            p.pop();m--;
        }
        res++;
    }
    writ(res); ptn;
}



int main() {
    //cin.tie(nullptr)->sync_with_stdio(false);
    //freopen("data.in", "r", stdin);freopen("data1.out", "w", stdout);

    int T = read(); while (T--)
        Qingtuan();
    WA
}