UVALive 5844 dfs暴力搜索

时间:2023-03-08 16:28:18
UVALive 5844 dfs暴力搜索

题目链接:UVAive 5844 Leet

DES:大意是给出两个字符串。第一个字符串里的字符可以由1-k个字符代替。问这两个字符串是不是相等。因为1<=k<=3。而且第一个字符串长度小于等于15.所以。对第一个字符串里的每一个字符尝试匹配1-k个字符看是否有可能相等就好了。

比赛的时候想到这是dfs类暴力,但是map<char, char*>没用过。不知道怎么记录了。而且dfs本身就不太会用。依然感觉dfs很奇妙。

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <map>
using namespace std; int k;
char s1[], s2[];
int len1, len2;
bool flag;
map<char, char *>mm; void dfs(int l1, int l2)
{
if (flag) return;
if (l1 == len1 && l2 == len2)
{
flag = true;
return;
}
if (l1 >= len1 || l2 >= len2) return;
if (mm[s1[l1]]) //如果这个字符以前出现过。像以前一样匹配。
{
char *temp = mm[s1[l1]]; // 这里用指针。不然不能直接赋值。
int lent = strlen(temp);
for (int i=, j=l2; i<lent; ++i, ++j) //是不是可以匹配。
{
if (temp[i] != s2[j])
return;
}
dfs(l1+, l2+lent); //继续匹配第一个字符串的下一个字符。
}
else //如果没出现过。匹配。
{
for (int i=; i<=k; ++i) //尝试把这个字符匹配给对应1-k个字符。k种。
{
char temp[]; //数组。用指针程序会崩。
memset(temp, , sizeof(temp));
for (int j=; j<i; ++j)
{
temp[j] = s2[l2+j];
}
mm[s1[l1]] = temp;
dfs(l1+, l2+i); //继续匹配下一个字符、
mm[s1[l1]] = ; //回溯。
}
}
} int main()
{
int t;
//ios::sync_with_stdio(false);
cin >> t;
//scanf("%d", &t);
while(t--)
{
//scanf("%d%s%s", &k, s1, s2);
cin >> k >> s1 >> s2;
flag = false;
mm.clear();
len1 = strlen(s1);
len2 = strlen(s2);
dfs(, );
if (flag) cout << << endl;
else cout << << endl;
}
return ;
}

(⊙o⊙)哦....学了一个新的东西....ios::sync_with_stdio(false);听说用了它..C++就可以接近C的时间啦。。。。。

试验了一下,,,,

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分别是用STL & C++ , C, C++...