24暑假集訓day2下午

Yantai_YZY發表於2024-08-03

下午

內容：STL 差分字首和倍增

1. STL

#include <iostream>
#include <queue>
#include <cmath>
#include <algorithm>
#include <vector>
#include <cstring>
#include <cstdio>
#include <set>
#include <map>
#include <unordered_map>
#include <bitset>
using namespace std;
void text(){
	vector<int>v;
	
	vector<int>A(5);
	
	vector<int>B(5,3);
        
	vector<int>C={1,2,3};
        
	for(vector<int>::iterator it=A.begin();it!=A.end();it++)cout<<*it<<" ";
	puts("");
        
	for(auto it:B)cout<<it<<" ";
	puts("");
	
	for(int i=0;i<(int)C.size();i++)cout<<C[i]<<" ";
	puts("");
      
	int n;
	cin >> n;
	v.resize(100);
	
	cout<<(int)v.size()<<"\n";//unsigned
	
	cout<<(int)v.capacity()<<"\n";
	
	for(int i=1;i<=n;i++)
		v.push_back(i);
	
	cout<<(int)v.size()<<"\n";//unsigned
	
	cout<<(int)v.capacity()<<"\n";
	
	v.shrink_to_fit();
	
	cout<<(int)v.capacity()<<"\n";
	
	v.push_back(2);
	
	cout<<(int)v.size()<<"\n";
	
	cout<<(int)v.capacity()<<"\n";
	
	v.clear();
	
	cout<<(int)v.capacity()<<"\n";
	
	vector<int>().swap(v);
	
	cout<<(int)v.capacity()<<"\n";
}

namespace basic
{
    basic_string<int>v;
    void test()
    {
        int n;
        cin >> n;
        v.resize(100);
        
        cout<<(int)v.size()<<"\n";//unsigned
        
        cout<<(int)v.capacity()<<"\n";
        
        for(int i=1;i<=n;i++)
            v.push_back(i);
        
        cout<<(int)v.size()<<"\n";//unsigned
        
        cout<<(int)v.capacity()<<"\n";
        
        v.shrink_to_fit();
        
        cout<<(int)v.capacity()<<"\n";
        
        v.push_back(2);
        
        cout<<(int)v.size()<<"\n";
        
        cout<<(int)v.capacity()<<"\n";
        
        v.clear();
        
        cout<<(int)v.capacity()<<"\n";
        
        basic_string<int>().swap(v);
        
        cout<<(int)v.capacity()<<"\n";
        
        basic_string<int>A={2,3,4};
        
        basic_string<int>B={2,3,4};
        
        for(auto it:A)cout<<it<<" ";
        puts("");
        
        A=A+B;
        
        for(auto it:A)cout<<it<<" ";
        puts("");
        
        //string 
        
        cout<<A.find({4,2})<<"\n"; 
        
        
    }
}
namespace se
{
    set<int>A;
    multiset<int>B;//set如果插入重複的元素會自動銷燬 , 而multiset不會
    void test()
    {
        for(int i=1;i<=3;i++)
        {
            A.insert(i);//pair <bool,iterator>
            A.insert(i);
            
            B.insert(i);//指標 iterator  
            B.insert(i);
        }
        for(auto it:A)cout<<it<<" ";
        puts("");
        
        for(auto it:B)cout<<it<<" ";
        puts("");
        
        multiset<int>::iterator x=B.lower_bound(0);
        
        multiset<int>::iterator y=B.find(2);
        
        B.erase(x);
        
        for(auto it:B)cout<<it<<" ";
        puts("");
        
        B.erase(3);
        
        for(auto it:B)cout<<it<<" ";
        puts("");
        
        cout<<(int)B.size()<<"\n";
        
        cout<<(int)B.count(2)<<"\n";
        
        A.clear(),B.clear();
        //不能陣列下標訪問。 
        
    }
}
namespace ma
{
    map<int,int>A;
    unordered_map<int,int>B;
    void test()
    {
        for(int i=1;i<=3;i++)A[i]=i-1,B[i]=i-1;
        
        for(auto it:A)cout<<it.first<<" "<<it.second<<"\n";
        puts("");
        
        auto it=A.find(2);//key 
        
        cout<<A[4]<<"\n";
        
        A.clear(),B.clear();
        
    }
}
namespace bit
{
    bitset<10>bit,bb;
    void test()
    {
        bit[5]=1;
        cout<<bit<<"\n";
        
        bit.flip();
        bit<<=3;
        cout<<bit<<"\n";
        
        bb[1]=1;
        bb[4]=1;
        
        cout<<bb<<"\n";
        bit^=bb;
        cout<<bit<<"\n";
        
        int A=bit._Find_first();
        cout<<A<<"\n";
        
        cout<<bit._Find_next(A)<<"\n";
        
        
    }
}
int main(){
	vector<int> a = {1,2,3};
	for(auto it:a){
		cout << it << " ";
	} 
	text();
	
	basic::test();
	basic_string<int>aa = {1, 2, 3};//標頭檔案在iostream 
	for(auto it:aa){
		cout << it << " ";
	} 
	
	return 0;
}

2. 差分字首和

不難，可以解決區間問題

T1 求區間和

std：

#include <iostream>
#include <queue>
#include <cmath>
#include <algorithm>
#include <vector>
#include <cstring>
#include <cstdio>
#include <set>
#include <map>
#include <unordered_map>
#include <bitset>
using namespace std;
const int MAXX = 1e5 + 10;
int a[MAXX], n; 
int main(){
	cin >> n;
	for(int i = 1;i <= n;i++){
		int x;
		cin >> x;
		a[i] = x + a[i - 1];
	}
	int q;
	cin >> q;
	while(q--){
		int l, r;
		cin >> l >> r;
		cout << a[r] - a[l - 1] << '\n';
	}
	return 0;
}

記錄

差分

實現：對於每個區間 \([l, r]\)，讓 \(a_l\) 加上 \(v\)，\(a_r+1\) 減去 \(v\)。

最大加權矩形

給定一個 \(n \cdot n\) 的矩形，詢問這個矩形內部和最大的一個子子矩形，這個子矩形的和是多少？
\(n ≤ 120\)

std：

const int xx=1005;
int n,q;
ll a[xx][xx];
int main(){
	n=read();
	for(int i=l;i<=n;i++)
		for(int j=1;j<=n;j++)
			a[i][j]=read();
	ll ma=0;
	for(int x=1;x<=n;x++)
		for(int y=1;y<=n;y++)
			for(int i=x;i<=n;i++)
				for(int j=y;j<=n;j++)
					ma=max(ma,a[i][j]+a[x-1][y-1]-a[i][y-1]-a[x-1][j]);
	cout<<ma<<"\n";
	return 0;
}

二維陣列字首和

T2 地毯

問題簡述：
在 \(n\times n\) 的格子上有 \(m\) 個地毯。
給出這些地毯的資訊，問每個點被多少個地毯覆蓋。

std：

#include <iostream>

#define maxn 1005

int n,m,a[maxn][maxn];

int main(){
    
    std::cin.tie(0)->sync_with_stdio(0);
    
    std::cin >> n >> m;
    for (int i = 1;i <= m;i++){
        
        int x1,y1,x2,y2;
        
        std::cin >> x1 >> y1 >> x2 >> y2;
        for (int j = x1;j <= x2;j++)
            for (int k = y1;k <= y2;k++)
                a[j][k]++;
        
    }
    for (int i = 1;i <= n;i++){
        for (int j = 1;j <= n;j++)
            std::cout << a[i][j] << ' ';
        std::cout << '\n';
        
    }
    
    return 0;
}

記錄

倍增（ST 表）

可解決的問題：區間極值

問題簡述：給定一個長度為 \(N\) 的數列，和 $ M $ 次詢問，求出每一次詢問的區間內數字的最大值。

std：

#include <iostream>
#include <queue>
#include <cmath>
#include <algorithm>
#include <vector>
#include <cstring>
#include <cstdio>
#include <set>
#include <map>
#include <unordered_map>
#include <bitset>
using namespace std;
inline int read(){
	int x=0,f=1;char ch=getchar();
	while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
	while (ch>='0'&&ch<='9'){x=x*10+ch-48;ch=getchar();}
	return x*f;
}
const int MAXN = 100005;
int n, m;
int a[MAXN], f[MAXN][21];
int main(){
	n = read(), m = read();
	for(int i = 1;i <= n;i++){
		a[i] = read();
		f[i][0] = a[i];
	}
	for(int j = 1;j <= 20;j++){
		for(int i = 1;i <= n;i++){
			if(i + (1 << j) - 1 <= n){
				f[i][j] = max(f[i][j - 1], f[i + (1 << (j - 1))][j - 1]);
			}
		}
	}
	while(m--){
		int l = read(), r = read();
		int k = __lg(r - l + 1);
		cout << max(f[l][k], f[r - (1 << k) + 1][k]) << '\n';
	}
	return 0;
}

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