STL _牛客博客

#include<bits/stdc++.h>
using namespace std;

map<int,int> mp;
set<int> st;

int main() {
    mp.insert({1,2});
    mp.insert({2,4});
    auto [a,b] = *mp.begin();
    auto [c,d] = *mp.rbegin();
    cout<<a<<" "<<b<<endl;
    cout<<c<<" "<<d<<endl;
    st.insert(1);
    st.insert(2);
    cout<<*st.begin()<<endl;
    cout<<*st.rbegin()<<endl;
    return 0;
}

multiset<int> st;

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
//set<int> st;
multiset<int> st;
int main(){
	st.insert(0);
	st.insert(2);
	st.insert(4);
	st.insert(10);
	st.insert(2);
	for(auto v : st) cout<<v<<" ";
	cout<<"\n";
	int x;
	cin >> x;
	cout<<*st.lower_bound(x)<<endl;
	if(st.lower_bound(x) == st.end()) cout<<"NO\n";
	else cout<<"YES\n";
	return 0;
}



若set不存在key则返回end迭代器: st.lower_bound(k) == end 判断条件！
map嵌套： map<int,map<int,int>> mp; mp[1][2] = 3;
set.lower_bound({pair}),set<tuple<int,int,int,int>> s;
map.lower_bound({pair});
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
const int mod = 1e9 + 7;
const int N = 1e6+9;
typedef long long ll;
typedef pair<int,int> PII;

map<int,int> mp;
unordered_set<int> st;
int a[N];

int main() {
    int n,q;
    cin >> n >> q;
    for(int i=1; i<=n; i++) {
        int x;
        scanf("%d",&x);
        if(!st.count(x)) {
            st.insert(x);
            mp[x] = i;
        }
    }
    while(q--) {
        int x;
        scanf("%d",&x);
        auto [a,b] = *mp.lower_bound(x);
        if(a == x) cout<<b<<" ";
        else cout<<"-1 ";
    }
    return 0;
}



https://www.luogu.com.cn/problem/P2249
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
const int mod = 1e9 + 7;
const int N = 1e6+9;
typedef long long ll;
typedef pair<int,int> PII;

unordered_map<int,int> mp;
set<PII> st;
set<tuple<int,int,int,int>> s;
int a[N];

int main() {
    int n,q;
    cin >> n >> q;
    for(int i=1; i<=n; i++) {
        int x;
        scanf("%d",&x);
        if(!mp[x]) {
            mp[x] = 1;
            st.insert({x,i});
        }
    }
    while(q--) {
        int x;
        scanf("%d",&x);
        auto [a,b] = *st.lower_bound({x,0});
        if(a == x) cout<<b<<" ";
        else cout<<"-1 ";
    }
    return 0;
}

刚刚看了好像deque也是O（1）插入和删除操作并且迭代器可以进行运算！ deque迭代器好像动态的,没弄明白不太好用!

List

list的erase(iterato'pos) 会返回下一个元素迭代器的位置，若在end-1(end 无数字)的位置会返回end迭代器即L.size() 其中只需要保证it的正确性即可！

STL 中有用于操作迭代器的三个函数模板，它们是： advance(p, n)：使迭代器 p 向前或向后移动 n 个元素。 distance(p, q)：计算两个迭代器之间的距离，即迭代器 p 经过多少次 + + 操作后和迭代器 q 相等。如果调用时 p 已经指向 q 的后面，则这个函数会陷入死循环。 iter_swap(p, q)：用于交换两个迭代器 p、q 指向的值。

//最终于版本!
#pragma comment(linker, "/STACK:102400000,102400000") 手动扩栈空间！

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

const int N = 1e5 + 9;
int a[N];
list<int> L;
int main()
{
    int n, k;
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; i++)
        scanf("%d", &a[i]), L.emplace_back(a[i]);
    int ans = 0;
    while (1)
    {
        int f = 0;
        for(auto it = L.begin(); it != L.end();)
        {
            if(L.size() <= 1) break;
            auto pre = it,ne = it;
            advance(pre,-1);
            advance(ne,1);
            if(it == L.begin()) pre = L.end(),pre--;
            auto x = L.end();
            if(it == --x) ne = L.begin();
            if(*it == *ne || *it + *ne == k)
            {
                f = 1;
                ans++;
                L.erase(it);
                it = L.erase(ne);
            }
            else if(*it == *pre || *it + *pre == k)
            {
                f = 1;
                ans++;
                L.erase(pre);
                it = L.erase(it);
            }
            else it++;
        }
        if(f == 0) break;
    }
    cout<<ans<<"\n";
    return 0;
}

https://ac.nowcoder.com/acm/contest/33192/F
因为没有特判1，导致RE了一个小时,呜呜~
#pragma comment(linker, "/STACK:102400000,102400000") 
#include<bits/stdc++.h>
using namespace std;

typedef long long ll;

const int N = 1e5+9;
int a[N];
list<int> L;
int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    for(int i=1; i<=n; i++) scanf("%d",&a[i]),L.emplace_back(a[i]);
    int ans = 0;
    int cnt = 0;
    while(1)
    {
        int f = 0;
        for(auto it = L.begin(); it != L.end();)
        {
            if(L.size() == 1) break;
            ++cnt;
            auto ne = it,pre = it;
            ne++,pre--;
            if(it == L.begin()) pre = L.end(),pre--;
            auto x = L.end();
            x--;
            if(it == x) ne = L.begin();
            if(*it == *ne || *it + *ne == k)
            {
                f = 1;
                if(ne == L.begin()) 
                {
                    L.erase(it);
                    it = L.end();
                    it--;
                    ne = L.erase(ne);
                }
                else 
                {
                    it = L.erase(it);
                    ne = L.erase(ne);
                    it = ne;
                }
                ans++;
            }
            else if(*it == *pre || *it + *pre == k)
            {
                if(it == L.begin()) 
                {
                    it = L.erase(it);
                    pre = L.erase(pre);
                    pre = L.end();
                    pre--;
                }
                else 
                {
                    pre = L.erase(pre);
                    it = L.erase(pre);
                    pre = it;
                }
                f = 1;
                ans++;
            }
            else it++;
        }
        if(f == 0) break;
    }
    printf("%d\n",ans);
    return 0;
}

/************
10 5
1 1 4 5 1 4 4 2 3 6
************/

#include<bits/stdc++.h>
using namespace std;

int main()
{
    list<int> L;
    int n;
    cin >> n;
    for(int i=1; i<=n; i++) L.push_back(i);
    L.push_front(0);
    auto del = L.begin();
    for(int i=1; i<=2; i++) del++;
    auto end = L.end();
    end--;
    for(auto it = del; it != end; it++)
    {
        auto x = it;
        x++;
        cout<<*it<<" "<<*x<<endl;
    }
    return 0;
}