POJ 1035

来源：互联网发布：《不持有的生活》知乎编辑：程序博客网时间：2024/05/01 11:28

从给出的字典中查找是否有符合的单词。如果没有这再次检查多一个字母或者少一个字母或者替换一个字母的单词，并把这些单词

按输入顺序全部输出出来。

#include <cstdio>#include <cstring>#include <algorithm>#include <vector>#include <string>#include <set>using namespace std;const int maxn = 10000 + 10;const int SIZE = 26;vector<int> ss;char s[maxn][SIZE], t[SIZE];int length[maxn];int n = 0;set<string> vis;bool Right(char * t){    return vis.count(string(t));}bool Insert(char *p, char *t, int len){    int cnt = 0, pos = -1;    for(int i = 0; i < len; ++i){        if(p[i] != t[i]) {            pos = i;            break;        }    }    if(pos == -1) return true;    for(int i = 0; i < pos; ++i){        if(p[i] != t[i]) return false;    }    for(int i = pos; i < len; ++i){        if(p[i + 1] != t[i]) return false;    }    return true;}bool Delete(char *p, char *t, int len){    return Insert(t, p, len - 1);}bool Replace(char *p, char *t, int len){    int cnt = 0;    for(int i = 0; i < len; ++i){        if(p[i] != t[i]) cnt ++;        if(cnt > 1) return false;    }    return true;}void Find(char *t, int len){    for(int i = 0; i < n; ++i){        if(length[i] == len && Replace(s[i], t, len)) ss.push_back(i);        if(length[i] == len - 1 && Delete(s[i], t, len)) ss.push_back(i);        if(length[i] == len + 1 && Insert(s[i], t, len)) ss.push_back(i);    }    sort(ss.begin(), ss.end());}int main(){    for(n = 0;;++n){        scanf("%s", s[n]);        if(strcmp(s[n], "#") == 0) break;        vis.insert(string(s[n]));        length[n] = strlen(s[n]);    }    for(;;){        scanf("%s", t);        if(strcmp(t, "#") == 0) break;        ss.clear();        if(Right(t)){            printf("%s is correct\n", t);        }else{            Find(t, strlen(t)); printf("%s:", t);            for(int i = 0; i < ss.size(); ++i){                printf(" %s", s[ss[i]]);            }printf("\n");        }    }    return 0;}

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