问题描述
问题要求编辑阶梯。编辑阶梯是一系列转换，其中编辑距离为 1，单词按字典顺序递增。根据onlinejudge board的想法，我实现了一个解决方案。首先生成一个无向图，其中每条边都意味着一个有效的转换/编辑步骤。然后使用 DFS 找到最大深度。但是我在以下测试用例中得到了错误的答案。
aaaaaa
aaaaaaa
aaaaaah
aaaaah
aaaah
aaah
aaarg
aaarge
aaarh
aabrge
aacrd
aacre
aaraah
aard
araah
ard
arde
arded
ardev
raah
rdev
vaah
vaahb
vaahe
答案应该是 11。但我的代码目前返回 8。代码如下所示。我正在做的唯一不同的步骤是在最后添加一个 $。这样做是为了避免我写两个循环。我有一个循环，我在当前字符之前添加字符，并且通过使用美元符号的虚拟字符，它也可以用于在字符串的末尾添加字符。我已经用单独的编辑步骤程序做了一些检查，生成的图形似乎是正确的。我很可能在 DFS 部分犯了一个错误。具体来说，这一行
if (!discovered[y] and ind_to_str[v] < ind_to_str[y])
对此的任何帮助表示赞赏。
#include <bits/stdc++.h>
#define MAX (25000 + 2)
using namespace std;
int max_depth = 0;
int dtime = 0;
vector<unordered_set<int>> graph;
vector<string> words;
unordered_map<string,int> str_to_ind;
unordered_map<int,string> ind_to_str;
vector<int> cur_chain,best_chain;
int parent[MAX];
int discovered[MAX];
int processed[MAX];
int entry_time[MAX];
int exit_time[MAX];
void dfs(int v,int d) {
cout << "At " << ind_to_str[v] << " with depth " << d << endl;
// max_depth = max(max_depth,d);
cur_chain.push_back(v);
if (d > max_depth) {
max_depth = d;
best_chain = cur_chain;
}
discovered[v] = true;
entry_time[v] = ++dtime;
for (int y : graph[v]) {
if (!discovered[y] and ind_to_str[v] < ind_to_str[y]) {
// if (!discovered[y]) {
parent[y] = v;
dfs(y,d + 1);
}
}
exit_time[v] = ++dtime;
processed[v] = true;
cur_chain.pop_back();
}
void init_search() {
for (int i = 0; i < MAX; ++i) {
parent[i] = entry_time[i] = exit_time[i] = -1;
discovered[i] = processed[i] = false;
}
dtime = 0;
}
int main(void) {
string line;
int ind = 0;
while (getline(cin,line) and line.length() > 0) {
// append an extra dollar so that you don't have to
// have an extra loop at the end for adding characters
// at the end of a string,you can have one loop which
// adds characters before every character and you've
// added characters in all places
line += "$";
words.push_back(line);
str_to_ind[line] = ind;
ind_to_str[ind++] = line;
}
int n = words.size();
graph.assign(n,unordered_set<int>());
for (int i = 0; i < n; ++i) {
for (int j = 0; j < words[i].length() - 1; ++j) {
for (char c = 'a'; c <= 'z'; ++c) {
// insert
string added = words[i].substr(0,j) + string(1,c) + words[i].substr(j + 1,string::npos);
if (str_to_ind.count(added) != 0) {
int cur_ind = str_to_ind[words[i]];
int next_ind = str_to_ind[added];
if (cur_ind == next_ind) continue;
graph[cur_ind].insert(next_ind);
graph[next_ind].insert(cur_ind);
}
// replace
string changed = words[i];
changed[j] = c;
if (str_to_ind.count(changed) != 0) {
int cur_ind = str_to_ind[words[i]];
int next_ind = str_to_ind[changed];
if (cur_ind == next_ind) continue;
graph[cur_ind].insert(next_ind);
graph[next_ind].insert(cur_ind);
}
}
// delete
string deleted = words[i].substr(0,j) + words[i].substr(j + 1,string::npos);
if (str_to_ind.count(deleted) != 0) {
int cur_ind = str_to_ind[words[i]];
int next_ind = str_to_ind[deleted];
if (cur_ind == next_ind) continue;
graph[cur_ind].insert(next_ind);
graph[next_ind].insert(cur_ind);
}
}
}
for (int i = 0; i < n; ++i) {
cout << ind_to_str[i] << ": ";
for (int x : graph[i]) {
cout << ind_to_str[x] << " ";
}
printf("\n");
}
for (int i = 0; i < n; ++i) {
cur_chain.clear();
init_search();
cout << "Starting with " << ind_to_str[i] << endl;
dfs(i,1);
}
printf("%d\n",max_depth);
for (int i = 0; i < best_chain.size(); ++i) {
cout << ind_to_str[best_chain[i]] << endl;
}
return 0;
}
解决方法
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