題意：

給你若干個(gè)字符串，答案串初始為空。第 $i$ 步將第 $i$ 個(gè)字符串加到答案串的后面，但是盡量地去掉重復(fù)部分（即去掉一個(gè)最長(zhǎng)的、是原答案串的后綴、也是第 $i$個(gè)串的前綴的字符串），求最后得到的字符串。
字符串個(gè)數(shù)不超過 $10^5$，總長(zhǎng)不超過 $10^6$

題目：

Amugae has a sentence consisting of n words. He want to compress this sentence into one word. Amugae doesn’t like repetitions, so when he merges two words into one word, he removes the longest prefix of the second word that coincides with a suffix of the first word. For example, he merges “sample” and “please” into “samplease”.

Amugae will merge his sentence left to right (i.e. first merge the first two words, then merge the result with the third word and so on). Write a program that prints the compressed word after the merging process ends.

Input

The first line contains an integer n (1≤n≤105), the number of the words in Amugae’s sentence.

The second line contains n words separated by single space. Each words is non-empty and consists of uppercase and lowercase English letters and digits (‘A’, ‘B’, …, ‘Z’, ‘a(chǎn)’, ‘b’, …, ‘z’, ‘0’, ‘1’, …, ‘9’). The total length of the words does not exceed 106.

Output

In the only line output the compressed word after the merging process ends as described in the problem.

Examples

Input

5
I want to order pizza

Output

Iwantorderpizza

Input

5
sample please ease in out

Output

sampleaseinout

分析：

KMP:從第二個(gè)開始字符串開始和前面合并的子串進(jìn)行kmp算法
注意每次匹配的起點(diǎn)位置，應(yīng)該為 文本串長(zhǎng)度-模式串長(zhǎng)度
哈希：每次需要求最長(zhǎng)的、是原答案串的后綴、也是第 $i$個(gè)串的前綴的字符串。枚舉這個(gè)串的長(zhǎng)度，哈希比較即可。

AC代碼：

KMP：

#include<bits/stdc++.h> using namespace std; const int M=1e6+10; string a,b; int nx[M]; int len,l2;void getnext() {int k = -1, j = 0;nx[0] = -1;while(j<l2){if(k==-1||b[j]==b[k])nx[++j]=++k;elsek=nx[k];} } int KMP(int st) {int i=st,j=0;getnext();while(i<len){if(j==-1 || a[i]==b[j]) ++i,++j;else j=nx[j];}return j; }int main() {int t;scanf("%d",&t);cin>>a;len=a.size();int pos=0;for(int i=1; i<t; i++){cin>>b;l2=b.size();int x=KMP(max(pos-l2,0));for(int i=x; i<l2; i++){a.push_back(b[i]);len++;}pos=len;}cout<<a<<endl; }

哈希

#include <bits/stdc++.h> using namespace std; typedef long long ll;const int N=1e6+10;ll mo[3]={(ll)1e9+21,(ll)1e9+9,(ll)1e9+7}; ll ba[3]={233,223,23}; ll p[3][N]; ll ha[3][N]; void init(){p[0][0]=p[1][0]=p[2][0]=1;for(int i=1; i<N; ++i){p[0][i]=(p[0][i-1]*ba[0])%mo[0];p[1][i]=(p[1][i-1]*ba[1])%mo[1];p[2][i]=(p[2][i-1]*ba[2])%mo[2];} } ll cal(int x,int l,int r){ll tp=(ha[x][r]-ha[x][l-1]*p[x][r-l+1])%mo[x];return (tp+mo[x])%mo[x]; }char c1[N],c2[N]; int getsr(int x,int n){int sr=1;ll d1=0,d2=0,d3=0;for(int i=n,j=1; i>=x; --i,++j){d1=(d1*ba[0]+c2[j])%mo[0];d2=(d2*ba[1]+c2[j])%mo[1];d3=(d3*ba[2]+c2[j])%mo[2];if(cal(0,i,n)==d1 && cal(1,i,n)==d2 && cal(2,i,n)==d3){sr=j+1;}}return sr; }int main() {init();int n;scanf("%d %s",&n,c1+1);int len=strlen(c1+1);for(int i=1; i<=len; ++i){ha[0][i]=(ha[0][i-1]*ba[0]+c1[i])%mo[0];ha[1][i]=(ha[1][i-1]*ba[1]+c1[i])%mo[1];ha[2][i]=(ha[2][i-1]*ba[2]+c1[i])%mo[2];}for(int i=1; i<n; ++i){scanf("%s",c2+1);int t1=strlen(c2+1);int sr=getsr(max(len-t1+1,1),len);for(int i=sr; i<=t1; ++i){++len;c1[len]=c2[i];ha[0][len]=(ha[0][len-1]*ba[0]+c1[len])%mo[0];ha[1][len]=(ha[1][len-1]*ba[1]+c1[len])%mo[1];ha[2][len]=(ha[2][len-1]*ba[2]+c1[len])%mo[2];}}printf("%s\n",c1+1);return 0; }

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