Kattis - secretchamber

拉什莫尔山不仅以其壮观的总统雕像闻名,还隐藏着一个鲜为人知的秘密室——记录厅,用于存放重要历史文件。本文揭秘了一位考古学家如何发现并试图解读隐藏在记录厅内的神秘文档。

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By now you have probably heard that there is a spectacular stone sculpture featuring four famous U.S. presidents at Mount Rushmore. However, very few people know that this monument contains a secret chamber. This sounds like something out of a plot of a Hollywood movie, but the chamber really exists. It can be found behind the head of Abraham Lincoln and was designed to serve as a Hall of Records to store important historical U.S. documents and artifacts. Historians claim that the construction of the hall was halted in 1939 and the uncompleted chamber was left untouched until the late 1990s, but this is not the whole truth.

In 1982, the famous archaeologist S. Dakota Jones secretly visited the monument and found that the chamber actually was completed, but it was kept confidential. This seemed suspicious and after some poking around, she found a hidden vault and some documents inside. Unfortunately, these documents did not make any sense and were all gibberish. She suspected that they had been written in a code, but she could not decipher them despite all her efforts.

Earlier this week when she was in the area to follow the ACM-ICPC World Finals, Dr. Jones finally discovered the key to deciphering the documents, in Connolly Hall of SDSM&T. She found a document that contains a list of translations of letters. Some letters may have more than one translation, and others may have no translation. By repeatedly applying some of these translations to individual letters in the gibberish documents, she might be able to decipher them to yield historical U.S. documents such as the Declaration of Independence and the Constitution. She needs your help.

You are given the possible translations of letters and a list of pairs of original and deciphered words. Your task is to verify whether the words in each pair match. Two words match if they have the same length and if each letter of the first word can be turned into the corresponding letter of the second word by using the available translations zero or more times.

Input

The first line of input contains two integers mm (1m5001≤m≤500) and nn (1n501≤n≤50), where mm is the number of translations of letters and nn is the number of word pairs. Each of the next mm lines contains two distinct space-separated letters aa and bb, indicating that the letter aa can be translated to the letter bb. Each ordered pair of letters (a,b)(a,b) appears at most once. Following this are nn lines, each containing a word pair to check. Translations and words use only lowercase letters ‘a’–‘z’, and each word contains at least 11 and at most 5050 letters.

Output

For each pair of words, display yes if the two words match, and no otherwise.

Sample Input 1Sample Output 1
9 5
c t
i r
k p
o c
r o
t e
t f
u h
w p
we we
can the
work people
it of
out the
yes
no
no
yes
yes

Sample Input 2

Sample Output 2

3 3 a c b a a b aaa abc abc aaa acm bcm

yes no yes


题意:
       前面给出n个字符,前面的字符可以转化为后面的字符,可以连续转化。m串字符,文两串字符是否相等。

思路:
       这题解法很多,可以用搜索,最短路,矩阵快速幂。搜索有一个坑点会内存超限或者时间超限,因为它会形成换环,a->b,b->c,c->a;这样会死循环,需要记录搜索过的点,然后不用再搜索。以下给出搜索和矩阵的解法。

搜索ac代码:

#include<string.h>
#include<stdio.h>
int q[123][505];
int qp[123];
char p1[50],p2[50];
char a[5],b[5];
int len1,len2;
int t[8000],tp;
int chazhao(int x,int y)
{
    if(x==y)
        return 1;
    else
    {
        for(int i=0; i<qp[x]; i++)
        {
            int gg=0;
            for(int k=0; k<tp; k++)
            {
                if(q[x][i]==t[k])
                {
                    gg=1;
                }
            }
            if(gg)
                continue;
            t[tp++]=q[x][i];
            if(chazhao(q[x][i],y))
                return 1;
        }
    }
    return 0;
}
int panding()
{
    int flag=1;
    for(int i=0; i<len1; i++)
    {
        int x=p1[i],y=p2[i];
        tp=0;
        t[tp++]=x;
        if(!chazhao(x,y))
        {
            flag=0;
            break;
        }
    }
    return flag;
}
int main()
{
    int n,m,x,y;
    while(~scanf("%d%d",&n,&m))
    {
        memset(qp,0,sizeof(qp));
        for(int i=0; i<n; i++)
        {
            scanf("%s%s",a,b);
            x=a[0],y=b[0];
            q[x][qp[x]]=y;
            qp[x]++;
        }
        for(int i=0; i<m; i++)
        {
            scanf("%s%s",p1,p2);
            len1=strlen(p1);
            len2=strlen(p2);
            if(len1!=len2)
            {
                printf("no\n");
                continue;
            }
            if(panding())
            {
                printf("yes\n");
            }
            else
            {
                printf("no\n");
            }
        }
    }
}

矩阵ac代码:
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
int a[30][30];
char x[5],y[5];
char p1[55],p2[55];
void jzc()
{
    int t=26;
    while(t--)
    {
        for(int i=0; i<26; i++)
        {
            a[i][i]=1;
            for(int j=0; j<26; j++)
            {
                for(int k=0; k<26; k++)
                {
                    a[i][j]+=a[i][k]*a[k][j];
                    if(a[i][j])
                        a[i][j]=1;
                }
            }
        }
    }
}
int main()
{
    int n,m,flag;
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=1; i<=n; i++)
        {
            scanf("%s%s",x,y);
            a[x[0]-'a'][y[0]-'a']=1;
        }
        jzc();
        for(int j=1; j<=m; j++)
        {
            flag=1;
            scanf("%s%s",p1,p2);
            int len1=strlen(p1);
            int len2=strlen(p2);
            if(len1!=len2)
                flag=0;
            for(int i=0; i<len1; i++)
            {
                if(!flag)
                    break;
                if(!a[p1[i]-'a'][p2[i]-'a'])
                {
                    flag=0;
                }
            }
            if(flag)
                printf("yes\n");
            else
                printf("no\n");
        }
    }
}

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