带通配符*和?的kmp
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Wildcard
TimeLimit: 2000/1000 MS (Java/Others) Memory Limit: 125536/65536 K(Java/Others)Total Submission(s): 563 Accepted Submission(s):135
Problem Description
When specifying file names (or paths) inDOS, Microsoft Windows and Unix-like operating systems,theasterisk character (“*") substitutes for any zero or morecharacters, and thequestion mark (“?")substitutes for any one character.
Now give you a text and a pattern, you should judge whether thepattern matches the text or not.
Now give you a text and a pattern, you should judge whether thepattern matches the text or not.
Input
There are several cases. For each case,only two lines. The first line contains a text, which contains onlylower letters. The last line contains a pattern, which consists oflower letters and the two wildcards (“*", "?").
The text is a non-empty string and its size is less than 100,000,so do the pattern.
We ensure the number of “?” and the number of “*” in the patternare both no more than 10.
The text is a non-empty string and its size is less than 100,000,so do the pattern.
We ensure the number of “?” and the number of “*” in the patternare both no more than 10.
Output
Output “YES” if the pattern matches thetext, otherwise “NO”.
Sample Input
abcdefa*c*f
Sample Output
YES
Source
2011 Multi-University Training Contest 7 - Host by ECNU
给两个长度不超过100000的字符串, 一个是带有通配符?和*的模式串, 问能否匹配. 通配符不超过10个.
HDU 3901 Wildcard
题目大意:
给两个长度不超过100000的字符串, 一个是带有通配符?和*的模式串, 问能否匹配. 通配符不超过10个.
这题弄了差不多一天= =···不过还算是有收获吧!
方法是我自己YY出来的,代码老长==···希望能有更好的方法,忘各位神牛指教!!
我先把字符串处理成s1-s2-s3这样,就是两头没有'?'和'*'(其中'*'直接忽视,计算'?'的数目,然后对文本串减去该数目的长度),一以方便下面的处理。
然后分别对子串进行kmp,找出所有匹配。
再然后对'?'两边进行合并,是匹配串都是s1*s2*s3这样,然后就可以进行贪心了。
不过具体的实现我觉得有挺多细节的···具体代码有详细注释:
#include <cstdio>#include <cstring>static const int maxn = 100100;//这些数组下标为0处用来表示数组长度//前面四个都是用来储存子串的起点or终点的static int start[25][maxn], end[25][maxn], s[25][maxn], e[25][maxn], temp1[maxn], temp2[maxn];//front和back表示当前匹配的子串前or后紧接着的'?'个数,然后会加到start or end数组中int front, back;int fail[maxn];void kmp(char* str, char* pat, int num){ int i, j, k; memset(fail, -1, sizeof(fail)); start[num][0] = 0; end[num][0] = 0; for( i = 1; pat[i]; ++i ) { for( k = fail[i-1]; k >= 0 && pat[i] - pat[k+1]; k = fail[k] ); if( pat[k+1] == pat[i] ) fail[i] = k + 1; } i = j = 0; while( str[i] && pat[j] ) { if( str[i] == pat[j] ) { ++i, ++j; if( pat[j] == '\0' ) { start[num][++start[num][0]] = i - j - front; end[num][++end[num][0]] = i - 1 + back; j = fail[j-1] + 1; } } else if( j == 0 ) ++i; else j = fail[j-1] + 1; } front = back = 0;}static char str[maxn], p[maxn], t[maxn];//st表示的是该间断点的状态,1代表'?',0是'*'static bool st[maxn];int main(){ int i, j, k, ll, rr, mid; int len, cnt1, cnt2; bool ok; while( scanf("%s %s", str, p) != EOF ) { ok = 1; front = back = 0; //先处理一下p的头尾,并相应的对str做出改变 //使之变成统一的s1-s2-s3-s4形式,两头没有'?'or'*',方便下面的处理 len = strlen(p); for( i = cnt1 = 0; i < len && (p[i] == '*' || p[i] == '?'); ++i ) if( p[i] == '?' ) ++cnt1; for( j = len-1, cnt2 = 0; j >= 0 && (p[j] == '*' || p[j] == '?'); --j ) if( p[j] == '?' ) ++cnt2; if( i > j ) { if( cnt1 <= strlen(str) ) printf("YES 0 0\n"); else printf("NO\n"); continue; } p[j+1] = '\0'; str[strlen(str)-cnt2] = '\0'; strcpy(p, p+i); strcpy(str, str+cnt1); //对每个子串求它的匹配 len = strlen(p); for( i = j = k = 0; i < len; ++i ) { if( p[i] == '*' ) { if( !j ) continue; t[j] = '\0'; st[k] = 0; kmp(str, t, k++); j = 0; } else if( p[i] == '?' ) { if( !j ) { //例子: aa*??b //??应该算到b中,就是后一个子串 front = 1; while( p[i+1] == '?' ) { ++front; ++i; } continue; } back = 1; while( p[i+1] == '?' ) { //例子: aa??? ++back; ++i; } //例子: aa???*b 与 aa???b st[k] = p[i+1] == '*' ? 0 : 1; t[j] = '\0'; kmp(str, t, k++); j = 0; } else t[j++] = p[i]; } t[j] = '\0'; kmp(str, t, k); //检查每个子串是否都出现了 for( i = 0; ok && i <= k; ++i ) if( !start[i][0] )ok = 0; //对状态为1,就是'?'两边的子串合并,使p串成为统一由'*'分隔的,以进行贪心 if( ok ) { len = 0; for( i = 0; ok && i < k; ++i ) { //临时数组,储存合并后的始点和终点 temp1[0] = temp2[0] = 0; if( st[i] ) { //枚举左边的子串的始点,然后对右边的终点进行二分查找 for( j = 1; j <= end[i][0]; ++j ) { ll = 1, rr = start[i+1][0]+1; while( rr - ll > 1 ) { mid = (ll+rr)/2; if( start[i+1][mid] <= end[i][j] + 1 ) ll = mid; else rr = mid; } if( start[i+1][ll] == end[i][j] + 1 ) { temp1[++temp1[0]] = start[i][j]; temp2[++temp2[0]] = end[i+1][ll]; } } //找不到符合的,匹配失败 if( !temp1[0] ) ok = 0; else { //把temp中的复制去i+1 memcpy(start[i+1], temp1, sizeof(temp1)); memcpy(end[i+1], temp2, sizeof(temp2)); } } else { for( s[len][0] = e[len][0] = 0, j = 1; j <= end[i][0]; ++j ) { //s, e储存合并后的子串 s[len][++s[len][0]] = start[i][j]; e[len][++e[len][0]] = end[i][j]; } ++len; } } //不要忘了最后一个 for( s[len][0] = e[len][0] = 0, j = 1; j <= end[k][0]; ++j ) { s[len][++s[len][0]] = start[k][j]; e[len][++e[len][0]] = end[k][j]; } ++len; } //贪心验证 if( ok ) { k = -1; for( i = 0; ok && i < len; ++i ) { for( j = 1; j <= s[i][0]; ++j ) if( s[i][j] > k ){k = e[i][j]; // 取第一种结果,即匹配的主串片断中长度最小的break;} if( j > s[i][0] ) ok = 0; } } if( ok ){ printf("(suffix = include ? : not *) \nYES\nfront = %d\nback = k + cnt1 + cnt2= %d\ns=\n", s[0][1], k + cnt1 + cnt2);for (int i = 0; i < len; i++){printf("%3d: ", s[i][0]);for(int j = 1; j <= s[i][0]; j++)printf("%3d ", s[i][j]);printf("\n");}printf("YES k = %d cnt1 = %d cnt2 = %d\ne=\n", k, cnt1, cnt2);for ( i = 0; i < len; i++){printf("%3d: ", e[i][0]);for(int j = 1; j <= e[i][0]; j++)printf("%3d ", e[i][j]);printf("\n");}} else printf("NO k = %d\n", k);printf("///////////////////////////////////////////////////////////\n"); } return 0;}
再给出几组我debug中比较有价值的数据:
abababcdababcdecdabefcda
*ab??cd??ef*
abcdebcdde
*abcd?e*
babbbabab
ab?b?bab
abcdef
a*b*c*d*e*f
isdjkasd
i*s*d*j*k*a*s*d
hellokugou
hello*??gou
dfjijijiugnmlok
??*f?ij*ug?ml?k
dfjijijiugnmlok
?*f?ij*ug?ml?k
abcdefghijklmnopqrstuvwx
ab*?*ef?h?jk*qr??*u??x?z
sodfmkkoasa
s?df?k?o?*a
sodfmkkoas
s?df?k?o?*
求更简单的方法···
http://blog.sina.com.cn/s/blog_7da04dd30100vlcp.html
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