正题

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题目大意

若干个字符串，每个字符串求一个前缀，使只有这个字符串有这个前缀。

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解题思路

$O(n^2)$枚举两个字符串，然后$O(n)$求出至少要取到哪里做前缀这两个字符串才不会冲突。

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代码

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;
int n,m,ms,ans,l[51];
char s[51][51];
int main()
{//freopen("abbreviate.in","r",stdin);//freopen("abbreviate.out","w",stdout);scanf("%d",&n);for (int i=1;i<=n;i++){scanf("%s",s[i]);l[i]=strlen(s[i]);//输入取长度}for (int i=1;i<=n;i++){ans=-1;for (int j=1;j<=n;j++){if (i==j) continue;ms=min(l[i],l[j]);for (int k=0;k<ms;k++){if (s[i][k]==s[j][k])//重复就需要继续往后取{ans=max(ans,k);//取最大值}else break;}}for (int k=0;k<=ans+1;k++)printf("%c",s[i][k]);//输出printf("\n");}
}

对拍程序

随机数据

#include<cstdio>
#include<ctime>
#include<cstdlib>
#include<string>
#include<iostream>
#include<map>
#define random(x) rand()%x+1
using namespace std;
map<string,bool> ok,ok2;
int n,m;
string s;
int main()
{freopen("abbreviate.in","w",stdout);srand((unsigned)time(0));n=random(50);printf("%d\n",n);for (int i=1;i<=n;i++){m=random(50);s="";int j;for (j=1;j<=m;j++){s+='a'+random(26)-1;if (ok2[s]) break;}if (!ok[s]&&j==m+1){ok2[s]=true;cout<<s;printf("\n");string ss="";for (int i=0;i<s.size();i++){ss+=s[i];ok[ss]=true;}}else n++;}
}

对拍

#include<cstdio>
#include<ctime>
#include<cstdlib>
#include<cstring>
using namespace std;
int main()
{for (int t=1;t<=100000;t++){system("abbreviatedata.exe");double st=clock();system("abbreviate.exe");double ed=clock();freopen("abbreviate.in","r",stdin);char s[51][51],d[51];int ans=0,n;scanf("%d",&n);for (int i=1;i<=n;i++){scanf("%s",s[i]);}fclose(stdin);freopen("abbreviate.out","r",stdin);for (int i=1;i<=n;i++){scanf("%s",d);int l=strlen(d);ans=0;for (int j=1;j<=n;j++){bool flag=false;for (int k=0;k<l;k++){if (s[j][k]!=d[k]){flag=true;break;}}if (!flag) {ans++;}}if (ans!=1) {printf("Wrong Answer ");printf("%d",i);return 0;}}fclose(stdin);printf("AC point:%d time:%.0lfms\n",t,ed-st);}
}