正題

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題目大意

若干個字符串，每個字符串求一個前綴，使只有這個字符串有這個前綴。

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解題思路

O(n2)O(n2)枚舉兩個字符串，然后O(n)O(n)求出至少要取到哪里做前綴這兩個字符串才不會沖突。

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代碼

#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]);//輸入取長度}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])//重復就需要繼續往后取{ans=max(ans,k);//取最大值}else break;}}for (int k=0;k<=ans+1;k++)printf("%c",s[i][k]);//輸出printf("\n");} }

對拍程序

隨機數據

#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++;} }

對拍

#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);} }

總結

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