SCU oj 4438:Censor
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Censor
frog is now a editor to censor so-called sensitive words (敏感词).
She has a long text p
. Her job is relatively simple -- just to find the first occurence of sensitive word w
and remove it.
frog repeats over and over again. Help her do the tedious work.
Input
The input consists of multiple tests. For each test:
The first line contains 1
string w. The second line contains 1 string p
.
(1≤length of w,p≤5?106
, w,p
consists of only lowercase letter)
Output
For each test, write 1
string which denotes the censored text.
Sample Input
abc
aaabcbc
b
bbb
abc
ab
Sample Output
a
ab
题意: 给你一个主串,递归删除模式串。
比如: T: abc S: aaabcbc
aaabcbc->aabc->a
非常巧妙的KMP,我们用一个栈记录当前的字符以及其在模式串匹配的位置,当位置等于模式串长度之后,将模式串长度的串出栈,从栈顶元素开始继续匹配主串.时间复杂度 O(n).
#include <stdio.h> #include <iostream> #include <string.h> #include <stack> #include <algorithm> using namespace std; const int N = 5000005; struct Node{ char c; int k; }; char w[N],t[N],ans[N]; int Next[N]; void get_next(char *p){ int len = strlen(p); int i=0,k=-1; Next[0] = -1; while(i<len){ if(k==-1||p[i]==p[k]){ i++,k++; Next[i] = k; } else k = Next[k]; } } void Kmp(char *s,char *p){ int len1 = strlen(s),len2 = strlen(p); int i=0,j=0,len; stack <Node> stk; while(!stk.empty()) stk.pop(); while(i<len1){ if(j==-1||s[i]==p[j]){ i++,j++; stk.push(Node{s[i-1],j}); }else { j=Next[j]; } if(j==len2){ len = len2; while(!stk.empty()&&len--) stk.pop(); if(stk.empty()) j = 0; else j = stk.top().k; } } int k = 0; while(!stk.empty()){ ans[k++] = stk.top().c; stk.pop(); } for(int i=k-1;i>=0;i--) printf("%c",ans[i]); printf("\n"); } int main(){ while(scanf("%s%s",w,t)!=EOF){ get_next(w); Kmp(t,w); } return 0; }
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