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  • CF1285D Dr. Evil Underscores(分治)

    l

    思路:

    dfs搜素即可,从最高位开始搜。
    如果该位的左儿子或右儿子只有一个,继续搜索有的;否则,两者取min

    代码:

    // Problem: CF1285D Dr. Evil Underscores
    // Contest: Luogu
    // URL: https://www.luogu.com.cn/problem/CF1285D
    // Memory Limit: 250 MB
    // Time Limit: 1000 ms
    // 
    // Powered by CP Editor (https://cpeditor.org)
    
    #include<bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    typedef unsigned long long ull;
    typedef pair<ll, ll>PLL;
    typedef pair<int, int>PII;
    typedef pair<double, double>PDD;
    #define I_int ll
    inline ll read()
    {
        ll 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 - '0';
            ch = getchar();
        }
        return x * f;
    }
    
    inline void out(ll x){
    	if (x < 0) x = ~x + 1, putchar('-');
    	if (x > 9) out(x / 10);
    	putchar(x % 10 + '0');
    }
    
    inline void write(ll x){
    	if (x < 0) x = ~x + 1, putchar('-');
    	if (x > 9) write(x / 10);
    	putchar(x % 10 + '0');
    	puts("");
    }
    
    #define read read()
    #define closeSync ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
    #define multiCase int T;cin>>T;for(int t=1;t<=T;t++)
    #define rep(i,a,b) for(int i=(a);i<=(b);i++)
    #define repp(i,a,b) for(int i=(a);i<(b);i++)
    #define per(i,a,b) for(int i=(a);i>=(b);i--)
    #define perr(i,a,b) for(int i=(a);i>(b);i--)
    ll ksm(ll a, ll b, ll p)
    {
        ll res = 1;
        while(b)
        {
            if(b & 1)res = res * a % p;
            a = a * a % p;
            b >>= 1;
        }
        return res;
    }
    const int inf = 0x3f3f3f3f;
    #define PI acos(-1)
    const int maxn=4e5+100;
    
    int n;
    vector<int>v;
    
    int dfs(vector<int>v,int k){
    	if(k<0||v.size()==0) return 0;
    	vector<int>p1,p2;
    	for(int i=0;i<v.size();i++){
    		int j=v[i];
    		if(j&(1<<k)) p1.push_back(j);
    		else p2.push_back(j);
    	}	
    	if(p1.size()==0) return dfs(p2,k-1);
    	if(p2.size()==0) return dfs(p1,k-1);
    	return (1<<k)+min(dfs(p1,k-1),dfs(p2,k-1));
    }
    
    int main(){
    	n=read;
    	rep(i,1,n) v.push_back(read);
    	cout<<dfs(v,30)<<"
    ";
    	return 0;
    } 
    
    
    
    
    
    
    
    
    
    
    
    
    
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  • 原文地址:https://www.cnblogs.com/OvOq/p/15042865.html
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