UVa 843

題目：有一些單詞都成的字典，還有一些加密過的單詞組成的句子（除字母只有空格和換行）；

            求出一種可行的解碼方案（字母鍵間構成雙射），沒有解全都輸出*。

分析：搜索、回溯、dfs。對應每個句子中的單詞在字典里搜索對應的匹配項。

            利用回溯法搜索，每次在整個字典中尋找匹配項，如果矛盾則返回；

            矛盾：對應字母映射不同，或被映射的字母已經被別的字母映射到；

說明：有是好久沒做題了:-(。

#include <stdio.h>#include <stdlib.h>#include <string.h>char dict_word[1001][20];int  dict_word_length[1001];char line_word[81][20];int  line_word_length[81];char maps[128];int search(int index, int line_size, int dict_size){if (index >= line_size) {return 1;}char save[128];for (int c = 0; c <= 'z'; ++ c) {save[c] = maps[c];}for (int i = 0; i < dict_size; ++ i) {if (line_word_length[index] == dict_word_length[i]) {int flag = 1;for (int j = 0; j < line_word_length[index]; ++ j) {if (!maps[line_word[index][j]] && !maps[dict_word[i][j]-'a']) {maps[line_word[index][j]] = dict_word[i][j];maps[dict_word[i][j]-'a'] = 1;}else if (maps[line_word[index][j]] == dict_word[i][j]) {continue;}else {flag = 0;break;}}if (flag && search(index+1, line_size, dict_size)) {return 1;}for (int c = 0; c <= 'z'; ++ c) {maps[c] = save[c];}}}return 0;}int main(){int  n;char lines[81];scanf("%d",&n);for (int i = 0; i < n; ++ i) {scanf("%s",dict_word[i]);dict_word_length[i] = strlen(dict_word[i]);}getchar();while (gets(lines)) {memset(maps, 0, sizeof(maps));int count = 0, word_index = 0;for (int i = 0; lines[i]; ++ i) {if (lines[i] >= 'a' && lines[i] <= 'z') {line_word[word_index][count ++] = lines[i];}else {line_word[word_index][count] = 0;if (count > 0) {line_word_length[word_index] = count;word_index ++;count = 0;}}}if (count > 0) {line_word_length[word_index] = count;word_index ++;}int flag = search(0, word_index, n);for (int i = 0; lines[i]; ++ i) {if (lines[i] >= 'a' && lines[i] <= 'z') {if (flag) {printf("%c",maps[lines[i]]);}else {printf("*");}}else {printf(" ");}}puts("");}return 0;}