题目链接：点击打开链接

题意：

给定3个字符串，进行拼接

重复的一段可以覆盖，问拼接后最小的长度（若一个串s1是s2的子串，则s1可以认为嵌入了s2内，不需要进行拼接

思路：

kmp搞一下。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <set>
using namespace std;
#define N 300005
char T[N];//从0开始存  
int f[N];//记录P的自我匹配  
void getFail(int len, char *P)    
{    
    int i = 0, j = -1;    
    f[0] = -1;    
    while(i != len)    
    {    
        if(j == -1 || P[i] == P[j])    
            f[++i] = ++j;    
        else    
            j = f[j];    
    }    
}  
int KMP(int *f2, char *S1, char *S2, int lens1, int lens2){   //f2是S2的失配数组 
	getFail(lens2, S2);
    int pos = 0, len = lens1, j = 0, i = 0;
	int gg = 0;
    while(i <= len)   
    {   
        while(j!=-1 && S1[i] != S2[j]) j = f2[j];   
        i++, j++;
		gg = max(j, gg);
        if(i == len) {
			pos = max(pos, j);  
		}
    }   
	if(gg == lens2)return gg;
    return pos;  //这样得到的是S1的尾部和S2的前缀的  最大匹配位置（在S2中的位置） 
}/* */  
char s[3][N/3];
char tmp[N];
int l[3];
int hehe(int a,int b,int c){
	int pos = KMP(f, s[a], s[b], l[a], l[b]);

	for(int i = 0; i < l[a]; i++)tmp[i] = s[a][i];
	int top = l[a];
	if(pos!=l[b])
	for(; pos<l[b]; pos++)tmp[top++] = s[b][pos];
	tmp[top] = 0;
	pos = KMP(f, tmp, s[c], top, l[c]);
	return top+(l[c]-pos);
}
int main(){
	int i, j, u, v;
	while(~scanf("%s",s[0])){
		scanf("%s",s[1]);
		scanf("%s",s[2]);
		for(i=0;i<3;i++)l[i]=strlen(s[i]);
		int ans = hehe(0,1,2);
		ans = min(ans, hehe(0,2,1));
		ans = min(ans, hehe(1,0,2));
		ans = min(ans, hehe(1,2,0));
		ans = min(ans, hehe(2,0,1));
		ans = min(ans, hehe(2,1,0));
		cout<<ans<<endl;
	}
	return 0;
}
/*
xufuzdlsjxmevrtessfbwlnzzclcqwevnnucxyvhngnxhcbdfwq
wlwobhnmmgtfolfaeckbrnnglylydxtgtvrlmeeszoiuatzzzxufuzdlsjxmevrt
brnnglylydxtgtvrlmeeszoiuatzzzx

syvncqmfhautvxudqdhggz
hrpxzeghsocjpicuixskfuzupytsgjsdiyb
ybcmnmnbpndbxlxbzhbfnqvwcffvrdhtickyqhupmcehls
aa
aa
aa
a
b
c
xab
abx
c
x
ab
ab

*/