用兩個map實現(xiàn)一一對應的判定。

Crypt Kicker II

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Time limit:	1sec.	Submitted:	122
Memory limit:	32M	Accepted:	33

Source: Waterloo ACM Programming Contest Oct 4, 1998

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A common but insecure method of encrypting text is to permute the letters of the alphabet. That is, in the text, each letter of the alphabet is consistently replaced by some other letter. So as to ensure that the encryption is reversible, no two letters are replaced by the same letter.

A common method of cryptanalysis is the known plaintext attack. In a known plaintext attack, the cryptanalist manages to have a known phrase or sentence encrypted by the enemy, and by observing the encrypted text then deduces the method of encoding.

Your task is to decrypt several encrypted lines of text, assuming that each line uses the same set of replacements, and that one of the lines of input is the encrypted form of the plaintext

the quick brown fox jumps over the lazy dog

Input

The input consists of several lines of input. Each line is encrypted as described above. Subsequent test cases are separated with a single blank line.

Output

Decrypt each line and print it to standard output. If there is more than one possible decryption, any one will do. If decryption is impossible, output a single line:

No solution.

The encrypted lines contain only lower case letters and spaces and do not exceed 80 characters in length. There are at most 100 input lines.

Separate output for subsequent cases with a single blank line.

Sample Input

vtz ud xnm xugm itr pyy jttk gmv xt otgm xt xnm puk ti xnm fprxq xnm ceuob lrtzv ita hegfd tsmr xnm ypwq ktj frtjrpgguvj otvxmdxd prm iev prmvx xnmq Sample Output now is the time for all good men to come to the aid of the party the quick brown fox jumps over the lazy dog programming contests are fun arent they


說明：
每組case給出幾行text。 某一行與"the quick brown fox jumps over the lazy dog" 字母一一對應。
找出來并將其余行破譯。
用兩個map分別建立 上述text與可能與之一一對應行之間的對應。
用來判斷是否有一個字母對應了兩個或者兩個不同的字母對應同一個的非法情況。

代碼如下：
#include<iostream>
#include<stdio.h>
#include<vector>
#include <map>
using namespace std;

int main() {
int t = 0;
vector<string> a;
string tmp;
string sta = "the quick brown fox jumps over the lazy dog";
while (getline(cin, tmp)) {
if (t)cout << endl;
t++;
a.clear();
map<char, char> m, opm;
bool flag2 = 1;
do {
a.push_back(tmp);
if (tmp.length() == sta.length() && flag2) {
m.clear();
opm.clear();
bool flag = 1;
for (int i = 0; i < tmp.length(); i++) {
if (sta[i] == ' ' && tmp[i] != ' ') {
flag = 0;
break;
}
}
if (flag) {
bool flag3 = 1;
for (int i = 0; i < tmp.length(); i++) {
if (m.find(tmp[i]) != m.end()) {
if (m[tmp[i]] != sta[i]) {
flag3 = 0;
break;
}
} else if (opm.find(sta[i]) != opm.end()) {
if (opm[sta[i]] != tmp[i]) {
flag3 = 0;
break;
}
} else {
m[tmp[i]] = sta[i];
opm[sta[i]] = tmp[i];
}
}
if (flag3)
flag2 = 0;
}
}
} while (getline(cin, tmp) && tmp != "");
if (flag2) {
cout << "No solution.\n";
continue;
}
for (int i = 0; i < a.size(); i++) {
for (int j = 0; j < a[i].length(); j++)
putchar(m[a[i][j]]);
cout << endl;
}
}
return 0;
}

轉載于:https://www.cnblogs.com/yangchenhao8/archive/2011/06/09/2076893.html

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