hdu 4300 Clairewd’s message(KMP)
2014-03-19 22:20
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题目链接:hdu 4300 Clairewd’s message
题目大意:题目意思比较难懂,给出两个字符串,第一个字符串为编码规则,比如说样例1,q位于第一个位置,表示说q的编码是a。然后给出另一个字符串s,要求添加尽量少得字符,使得f可以被分成两部分f1+f2, 然后f1编码后等于f2,然后注意给出的f串为编码串,并非原串。
解题思路:首先先将f串转换成原串s,然后求出jump数组,去和f匹配,对于匹配的最后一项值p(即为最长匹配后缀),n-p即是我们要求的f1的长度(但是这里要注意n-p 不能小于p,因为分成两部分后不能说有公共的部分存在,所以p = jump[p])
#include <stdio.h>
#include <string.h>
const int N = 1e5+5;
int n, jump
;
char t[30], v[30], s
, f
;
void getJump () {
int p = 0;
for (int i = 2; i <= n; i++) {
while (p > 0 && s[p+1] != s[i])
p = jump[p];
if (s[p+1] == s[i])
p++;
jump[i] = p;
}
}
int KMP () {
int p = 0;
for (int i = 1; i <= n; i++) {
while (p > 0 && s[p+1] != f[i])
p = jump[p];
if (s[p+1] == f[i])
p++;
if (p == n)
p = jump[p];
}
while (n - p < p)
p = jump[p];
return p;
}
void init () {
scanf("%s%s", t, f+1);
n = strlen (f+1);
for (int i = 0; i < 26; i++)
v[t[i]-'a'] = i + 'a';
for (int i = 1; i <= n; i++)
s[i] = v[f[i]-'a'];
s[n+1] = '\0';
}
int main () {
int cas;
scanf("%d", &cas);
while (cas--) {
init ();
getJump ();
int k = n - KMP ();
for (int i = 1; i <= k; i++)
printf("%c", f[i]);
for (int i = 1; i <= k; i++)
printf("%c", v[f[i]-'a']);
printf("\n");
}
return 0;
}
题目大意:题目意思比较难懂,给出两个字符串,第一个字符串为编码规则,比如说样例1,q位于第一个位置,表示说q的编码是a。然后给出另一个字符串s,要求添加尽量少得字符,使得f可以被分成两部分f1+f2, 然后f1编码后等于f2,然后注意给出的f串为编码串,并非原串。
解题思路:首先先将f串转换成原串s,然后求出jump数组,去和f匹配,对于匹配的最后一项值p(即为最长匹配后缀),n-p即是我们要求的f1的长度(但是这里要注意n-p 不能小于p,因为分成两部分后不能说有公共的部分存在,所以p = jump[p])
#include <stdio.h>
#include <string.h>
const int N = 1e5+5;
int n, jump
;
char t[30], v[30], s
, f
;
void getJump () {
int p = 0;
for (int i = 2; i <= n; i++) {
while (p > 0 && s[p+1] != s[i])
p = jump[p];
if (s[p+1] == s[i])
p++;
jump[i] = p;
}
}
int KMP () {
int p = 0;
for (int i = 1; i <= n; i++) {
while (p > 0 && s[p+1] != f[i])
p = jump[p];
if (s[p+1] == f[i])
p++;
if (p == n)
p = jump[p];
}
while (n - p < p)
p = jump[p];
return p;
}
void init () {
scanf("%s%s", t, f+1);
n = strlen (f+1);
for (int i = 0; i < 26; i++)
v[t[i]-'a'] = i + 'a';
for (int i = 1; i <= n; i++)
s[i] = v[f[i]-'a'];
s[n+1] = '\0';
}
int main () {
int cas;
scanf("%d", &cas);
while (cas--) {
init ();
getJump ();
int k = n - KMP ();
for (int i = 1; i <= k; i++)
printf("%c", f[i]);
for (int i = 1; i <= k; i++)
printf("%c", v[f[i]-'a']);
printf("\n");
}
return 0;
}
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