UVA10150 POJ2647 Doublets【BFS】

Doublets
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 1090 Accepted: 371 Special Judge

Description

A Doublet is a pair of words that differ in exactly one letter; for example, “booster” and “rooster” or “rooster” and “roaster” or “roaster” and “roasted”.
You are given a dictionary of up to 25143 lower case words, not exceeding 16 letters each. You are then given a number of pairs of words. For each pair of words, find the shortest sequence of words that begins with the first word and ends with the second, such that each pair of adjacent words is a doublet. For example, if you were given the input pair “booster” and “roasted”, a possible solution would be: (“booster”, “rooster”, “roaster”, “roasted”) provided that these words are all in the dictionary.

Input

Input consists of the dictionary followed by a number of word pairs. The dictionary consists of a number of words, one per line, and is terminated by an empty line. The pairs of input words follow; the words of each pair occur on a line separated by a space.

Output

For each input pair, print a set of lines starting with the first word and ending with the last. Each pair of adjacent lines must be a doublet. If there are several minimal solutions, any one will do. If there is no solution, print a line: “No solution.” Leave a blank line between cases.

Sample Input

booster
rooster
roaster
coasted
roasted
coastal
postal

booster roasted
coastal postal

Sample Output

booster
rooster
roaster
roasted

No solution.

Source
Waterloo local 1999.01.31

问题链接：UVA10150 POJ2647 Doublets
问题简述：定义两个单词为doublet序列，当两个序列的长度相同，且只有一个字母不相同，给出一个单词库，计算两个单词之间可以通过doublet序列达到的路径。
问题分析：用BFS实现。
程序说明：代码在POJ中出现了TLE，程序需要改进。程序代码中使用了goto语言来跳出多重循环。
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* UVA10150 POJ2647 Doublets */#include <iostream>
#include <vector>
#include <map>
#include <queue>
#include <cstring>using namespace std;vector<string> dict;
map<string, int> pos;
const int N = 25143;
int next2[N];
string s, t;void bfs()
{memset(next2, -1, sizeof(next2));int ids = pos[s];int idt = pos[t];next2[idt] = -2;queue<int> q;q.push(idt);while(!q.empty()) {int u = q.front();q.pop();string s2 = dict[u];for(int i = 0; s2[i]; i++)for(int c = 'a'; c <= 'z'; c++) {string t2 = s2;t2[i] = c;if(t2 != s2) {if(pos.find(t2) != pos.end()) {int v = pos[t2];if(next2[v] == -1) {next2[v] = u;if(v == ids) goto stop;q.push(v);}}}}}stop:if(next2[ids] == -1) cout << "No solution." << endl;elsefor(int i = ids; i != -2; i = next2[i]) cout << dict[i] << endl;
}int main()
{int cnt = 0;string word;while(getline(cin, word) && word != "" ) {pos[word] = cnt++;dict.push_back(word);}int flag = 0;while(cin >> s >> t) {if(flag) cout << endl;else flag = 1;if(pos.find(s) == pos.end()) cout << "No solution." << endl;else if(pos.find(t) == pos.end()) cout << "No solution." << endl;else bfs();}return 0;
}

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