Sample Input
i
is
has
have
be
my
more
contest
me
too
if
award
#
me
aware
m
contest
hav
oo
or
i
fi
mre
#
Sample Output
me is correct
aware: award
m: i my me
contest is correct
hav: has have
oo: too
or:
i is correct
fi: i
mre: more me
这道题首先看到题目数据范围1e4,所以可以加剪枝暴力求解。
重点是要分析出长度相差1的和长度相等的才可以经过变换的到字典中的串。
在此基础上,查看匹配情况,长串作为被匹配的, 相同的字符ans++。
分情况分析是否能够变换得到。
最后注意输出格式和要求。
#include <cstdio> #include <cstring> #include <cmath> #include <cctype> #include <iostream> #include <algorithm> #include <map> #include <set> #include <vector> #include <string> #include <stack> #include <queue> typedef long long LL; typedef unsigned long long ULL; using namespace std; // bool Sqrt(LL n) { return (LL)sqrt(n) * sqrt(n) == n; } const double PI = acos(-1.0), ESP = 1e-10; const LL INF = 99999999999999; const int inf = 999999999, maxn = 1e4 + 24; string m, a; // vector<int> len1, len2; vector<string> v1, v2; int main() { //freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); while(cin >> m) { if(m == "#") break; v1.push_back(m); //len1.push_back(m.size()); } while(cin >> a) { if(a == "#") break; v2.push_back(a); //len2.push_back(a.size()); } int x = v1.size(), y = v2.size(); for(int i = 0; i < y; i++) { // if(find(v1.begin(), v1.end(), v2[i]) - v1.end() != v1.size() ) cout << v2[i] << " is correct"; bool flag = 0; for(int k = 0; k < x; k++) { if(v1[k] == v2[i]) { flag = 1; break;} } if(flag) { cout << v2[i] << " is correct "; continue; } else { cout << v2[i] << ":"; int l = v2[i].size(); for(int j = 0; j < x; j++) { int h = v1[j].size(); int ans = 0; if(l == h + 1) { for(int m = 0, d = 0; m < l; m++) if(v1[j][d] == v2[i][m]) { ans++; d++; } } else if(l + 1 == h) { for(int m = 0, d = 0; m < h; m++) if(v1[j][m] == v2[i][d]) { ans++; d++; } } else if(l == h) { for(int m = 0, d = 0; m < l; m++, d++) if(v1[j][d] == v2[i][m]) ans++; } if((l >= h) && (ans == l - 1)) cout << " " << v1[j]; if((l < h) && (ans == l)) cout << " " << v1[j]; } } puts(""); } return 0; } /* input: output: modeling: methods: complexity: summary: */