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  • Ural 1238 Folding 题解

    Ural 1238 Folding 题解

    题意

    定义折叠、展开为:

    • 单个大写英文字母是一个折叠的串,把它展开后是它本身。
    • 如果(S)(Q)是折叠的串,则(SQ)也是折叠的串。如果(S)展开后为(S')(Q)展开后为(Q'),则(SQ)展开后为(S'Q')
    • 如果(S)是个折叠的串,则(X(S))也是折叠的串,其中(X)是一个十进制大于(1)的整数,如果(S)展开为(S'),则(X(S))展开后为(S')重复(X)次。

    给定一个字符串(长度小于等于(100)),求把它折叠后有最小长度的那个字符串。

    题解

    考虑记忆化搜索(DP也可以)。

    定义(f(s))(s)字符串折叠后有最小长度的那个字符串。

    边界为当(|s|le4)((|s|)(s)的长度)时,(f(s)=s)

    给定(s),求出(f(s))有以下几种方式

    • (f(s)=s)

    • (1<xle|s|,|s|\%x=0)且有字符串(q)(q)重复(x)次为(s)(f(s)=x"("+f(q)+")")

    • (0<i<|s|)(f(s)=f(s[0..i-1])+f(s[i..|s|-1]))

    取长度最小的值即可,不要忘了记忆化。

    程序

    // #pragma GCC optimize(2)
    // #pragma G++ optimize(2)
    // #pragma comment(linker,"/STACK:102400000,102400000")
    
    // #include <bits/stdc++.h>
    #include <map>
    #include <set>
    #include <list>
    #include <array>
    #include <cfenv>
    #include <cmath>
    #include <ctime>
    #include <deque>
    #include <mutex>
    #include <queue>
    #include <ratio>
    #include <regex>
    #include <stack>
    #include <tuple>
    #include <atomic>
    #include <bitset>
    #include <cctype>
    #include <cerrno>
    #include <cfloat>
    #include <chrono>
    #include <cstdio>
    #include <cwchar>
    #include <future>
    #include <limits>
    #include <locale>
    #include <memory>
    #include <random>
    #include <string>
    #include <thread>
    #include <vector>
    #include <cassert>
    #include <climits>
    #include <clocale>
    #include <complex>
    #include <csetjmp>
    #include <csignal>
    #include <cstdarg>
    #include <cstddef>
    #include <cstdint>
    #include <cstdlib>
    #include <cstring>
    #include <ctgmath>
    #include <cwctype>
    #include <fstream>
    #include <iomanip>
    #include <numeric>
    #include <sstream>
    #include <ccomplex>
    #include <cstdbool>
    #include <iostream>
    #include <typeinfo>
    #include <valarray>
    #include <algorithm>
    #include <cinttypes>
    #include <cstdalign>
    #include <stdexcept>
    #include <typeindex>
    #include <functional>
    #include <forward_list>
    #include <system_error>
    #include <unordered_map>
    #include <unordered_set>
    #include <scoped_allocator>
    #include <condition_variable>
    // #include <conio.h>
    // #include <windows.h>
    using namespace std;
    
    typedef long long LL;
    typedef unsigned int ui;
    typedef unsigned long long ull;
    typedef float fl;
    typedef double ld;
    typedef long double LD;
    typedef pair<int,int> pii;
    #if (WIN32) || (WIN64) || (__WIN32) || (__WIN64) || (_WIN32) || (_WIN64) || (WINDOWS)
    #define lld "%I64d"
    #define llu "%I64u"
    #else
    #define lld "%lld"
    #define llu "%llu"
    #endif
    #define ui(n) ((unsigned int)(n))
    #define LL(n) ((long long)(n))
    #define ull(n) ((unsigned long long)(n))
    #define fl(n) ((float)(n))
    #define ld(n) ((double)(n))
    #define LD(n) ((long double)(n))
    #define char(n) ((char)(n))
    #define Bool(n) ((bool)(n))
    #define fixpoint(n) fixed<<setprecision(n)
    
    const int INF=1061109567;
    const int NINF=-1044266559;
    const LL LINF=4557430888798830399;
    const ld eps=1e-15;
    #define MOD (1000000007)
    #define PI (3.1415926535897932384626433832795028841971)
    
    /*
    #define MB_LEN_MAX 5
    #define SHRT_MIN (-32768)
    #define SHRT_MAX 32767
    #define USHRT_MAX 0xffffU
    #define INT_MIN (-2147483647 - 1)
    #define INT_MAX 2147483647
    #define UINT_MAX 0xffffffffU
    #define LONG_MIN (-2147483647L - 1)
    #define LONG_MAX 2147483647L
    #define ULONG_MAX 0xffffffffUL
    #define LLONG_MAX 9223372036854775807ll
    #define LLONG_MIN (-9223372036854775807ll - 1)
    #define ULLONG_MAX 0xffffffffffffffffull
    */
    
    #define MP make_pair
    #define MT make_tuple
    #define All(a) (a).begin(),(a).end()
    #define pall(a) (a).rbegin(),(a).rend()
    #define log2(x) log(x)/log(2)
    #define Log(x,y) log(x)/log(y)
    #define SZ(a) ((int)(a).size())
    #define rep(i,n) for(int i=0;i<((int)(n));i++)
    #define rep1(i,n) for(int i=1;i<=((int)(n));i++)
    #define repa(i,a,n) for(int i=((int)(a));i<((int)(n));i++)
    #define repa1(i,a,n) for(int i=((int)(a));i<=((int)(n));i++)
    #define repd(i,n) for(int i=((int)(n))-1;i>=0;i--)
    #define repd1(i,n) for(int i=((int)(n));i>=1;i--)
    #define repda(i,n,a) for(int i=((int)(n));i>((int)(a));i--)
    #define repda1(i,n,a) for(int i=((int)(n));i>=((int)(a));i--)
    #define FOR(i,a,n,step) for(int i=((int)(a));i<((int)(n));i+=((int)(step)))
    #define repv(itr,v) for(__typeof((v).begin()) itr=(v).begin();itr!=(v).end();itr++)
    #define repV(i,v) for(auto i:v)
    #define repE(i,v) for(auto &i:v)
    #define MS(x,y) memset(x,y,sizeof(x))
    #define MC(x) MS(x,0)
    #define MINF(x) MS(x,63)
    #define MCP(x,y) memcpy(x,y,sizeof(y))
    #define sqr(x) ((x)*(x))
    #define UN(v) sort(All(v)),v.erase(unique(All(v)),v.end())
    #define filein(x) freopen(x,"r",stdin)
    #define fileout(x) freopen(x,"w",stdout)
    #define fileio(x)
    	freopen(x".in","r",stdin);
    	freopen(x".out","w",stdout)
    #define filein2(filename,name) ifstream name(filename,ios::in)
    #define fileout2(filename,name) ofstream name(filename,ios::out)
    #define file(filename,name) fstream name(filename,ios::in|ios::out)
    #define Pause system("pause")
    #define Cls system("cls")
    #define fs first
    #define sc second
    #define PC(x) putchar(x)
    #define GC(x) x=getchar()
    #define Endl PC('
    ')
    #define SF scanf
    #define PF printf
    
    inline int Read()
    {
        int X=0,w=0;char ch=0;while(!isdigit(ch)){w|=ch=='-';ch=getchar();}while(isdigit(ch))X=(X<<3)+(X<<1)+(ch^48),ch=getchar();
    	return w?-X:X;
    }
    inline void Write(int x){if(x<0)putchar('-'),x=-x;if(x>9)Write(x/10);putchar(x%10+'0');}
    
    inline LL powmod(LL a,LL b){LL RES=1;a%=MOD;assert(b>=0);for(;b;b>>=1){if(b&1)RES=RES*a%MOD;a=a*a%MOD;}return RES%MOD;}
    inline LL gcdll(LL a,LL b){return b?gcdll(b,a%b):a;}
    const int dx[]={0,1,0,-1,1,-1,-1,1};
    const int dy[]={1,0,-1,0,-1,-1,1,1};
    /************************************************************Begin************************************************************/
    const int maxn=110;
    
    inline string Itoa(int x)
    {
    	string res="";
    	while(x)
    	{
    		res+=char(x%10+'0');
    		x/=10;
    	}
    	reverse(All(res));
    	return res;
    }
    
    map<string,string> vis;
    
    inline string sol(string s)
    {
    	if(vis.count(s)) return vis[s];
    	if(s.size()<=4) return vis[s]=s;
    
    	string ans=s;
    	rep1(i,s.size()) if(i>1&&s.size()%i==0)
    	{
    		int len=s.size()/i;
    		string x=s.substr(0,len);
    		bool f=1;
    
    		for(int j=0;j<s.size();j+=len) if(s.substr(j,len)!=x) f=0;
    
    		if(f)
    		{
    			string res=Itoa(i)+'('+sol(x)+')';
    			if(ans.size()>res.size()) ans=res;
    		}
    	}
    
    	rep1(i,s.size()-1)
    	{
    		string res=sol(s.substr(0,i))+sol(s.substr(i));
    		if(ans.size()>res.size()) ans=res;
    	}
    
    	return vis[s]=ans;
    }
    
    int main()
    {
    	string s;
    	cin>>s;
    	cout<<sol(s);
    
    	return 0;
    }
    /*************************************************************End**************************************************************/
    
    如有笔误或错误欢迎指出,谢谢。
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  • 原文地址:https://www.cnblogs.com/bitstd/p/11294451.html
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