如题，样例已过，TLE on test3，且当N = 1e6+10时无法运行，思路与第一篇题解雷同，已取公共最长公共前后缀的长度=min(字符串1的长度,字符串2的长度)，但仍TLE，求调

#include <bits/stdc++.h>
using namespace std;

const int N = 1e5 + 10;
//ne[j]:代表p字符串[1,j]这个字串前缀和后缀相等的最长长度
int ne[N]; 

int kmp(char s[],char p[]){
	ne[0] = ne[1] = 0;
	int lens = strlen(s+1),lenp = strlen(p+1);
	//计算字符串p的next的值
	for(int i=1,j=0;i < lenp;i++){
		while(j && p[i+1]!=p[j+1]) j = ne[j];
		if(p[i+1] == p[j+1]) j++;
		ne[i+1] = j;
	} 
	
	//KMP匹配
	for(int i = 0,j = 0;i < lens;i++){
		while(j && s[i+1] != p[j+1]) j = ne[j];
		if(s[i+1] == p[j+1]) j++;
		//判断如果匹配成功
		if(i+1==lens){//j == lenp
			return j+1;
		} 
	} 
}

int main(){ 
	int n;
	cin>>n;
	char c[N],ans[N],newans[N];
	scanf("%s",ans+1);
	int anslen=strlen(ans+1);
	for(int i=2;i<=n;i++){
		scanf("%s",c+1);
		int len=strlen(c+1);
		int minlen=min(strlen(ans+1),strlen(c+1));
//		newans[N]="";
		for(int i=1;i<=minlen;i++){
			newans[i]=ans[anslen-minlen+i];
		}
//		cout<<anslen<<" "<<minlen<<" ";
//		printf("%s %s\n",ans+1,newans+1);
		int x=kmp(newans,c);
//		printf("%s %s\n",ans+1,c+1);
//		cout<<x<<" "<<len<<endl;
		for(int j=x;j<=len;j++){
			ans[++anslen]=c[j];
		}
//		printf("%s\n",ans);
	}
	printf("%s",ans+1);
	return 0; 
}