scu 4438 Censor

之前做过类似的题目，不过因为匹配串长度较短，所以都是直接模拟的，而此题字符串的长度比较长，所以我们得用KMP。
先用KMP得到匹配串w的next数组，然后匹配两个串，边匹配边用栈模拟，栈维护主串每次匹配到匹配串的哪个位置，然后如果匹配到了整个串，那么将栈顶的 |匹配串 |全部删除，然后使得匹配串的位置移动到当初记录的位置，继续与主串进行匹配。

#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<iomanip>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>#include<map>#include<set>#include<bitset>//#define ONLINE_JUDGE#define eps 1e-5#define INF 0x7fffffff#define FOR(i,a) for((i)=0;i<(a);(i)++)#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,num) scanf("%d%d%d",&a,&b,&num)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))#define ll long longconst double PI=acos(-1.0);template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}template<class T> inline T Min(T a,T b){return a<b?a:b;}template<class T> inline T Max(T a,T b){return a>b?a:b;}using namespace std;//#pragma comment(linker,"/STACK:1024000000,1024000000")int n,m,x;#define N 5000010#define M 100010#define Mod 1000000000#define p(x,y) make_pair(x,y)const int MAX_len=550;char w[N];char p[N];int next[N];struct Node{    char c;    int pos;};void getNext(){    next[0]=-1;    int j=0,k=-1;    int len = strlen(w);    while(j<len){        if(k == -1 || w[j] == w[k]){            k++;            j++;            if(w[j] != w[k])                next[j] = k;            else                next[j] = next[k];        }else            k = next[k];    }}void KMP(){    int len1 = strlen(p);    int len2 = strlen(w);    if(len1<len2){        printf("%s\n",p);        return;    }    stack<Node> s;    int i=0,j=0;    while(i<len1){        if(j == -1 || p[i] == w[j]){            j++;            i++;            s.push((Node){p[i-1],j});//记录下该位置的字符和其匹配到的位置        }else            j = next[j];        if(j>=len2){            int tmp = len2;            while(tmp--) s.pop();   //将栈顶的len2个字符全部删除            if(s.empty()) j=0;  //如果栈空了，那么匹配串重新匹配            else                j = s.top().pos;//否则主串将继续与匹配串的pos位置匹配        }    }    vector<char> v;    while(!s.empty()){        v.push_back(s.top().c);        s.pop();    }    int size = (int)v.size();    for(int i=size-1;i>=0;i--)        printf("%c",v[i]);    printf("\n");}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);//  freopen("out.txt", "w", stdout);#endif    while(scanf("%s%s",w,p)!=EOF){        getNext();        KMP();    }    return 0;}