CodeForces 25E Test KMP

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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 300005char 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;}/*xufuzdlsjxmevrtessfbwlnzzclcqwevnnucxyvhngnxhcbdfwqwlwobhnmmgtfolfaeckbrnnglylydxtgtvrlmeeszoiuatzzzxufuzdlsjxmevrtbrnnglylydxtgtvrlmeeszoiuatzzzxsyvncqmfhautvxudqdhggzhrpxzeghsocjpicuixskfuzupytsgjsdiybybcmnmnbpndbxlxbzhbfnqvwcffvrdhtickyqhupmcehlsaaaaaaabcxababxcxabab*/

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