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  • POJ3080 Blue Jeans KMP+枚举

      题目链接:http://poj.org/problem?id=3080

      每个字符串的长度为60,而且字符串的数量很少,容易想到枚举水过。这题目的数据,就算纯暴力不优化,貌似也能过。加个KMP匹配也看不出多少优势。。。我在写KMP的时候,犯了一个低级的错误,居然把匹配过程写错了,导致wa了很久,下次吸取教训!!!

     1 //STATUS:C++_AC_0MS_164KB
     2 #include<stdio.h>
     3 #include<stdlib.h>
     4 #include<string.h>
     5 #include<math.h>
     6 #include<iostream>
     7 #include<string>
     8 #include<algorithm>
     9 #include<vector>
    10 #include<queue>
    11 #include<stack>
    12 #include<map>
    13 using namespace std;
    14 #define LL __int64
    15 #define pii pair<int,int>
    16 #define Max(a,b) ((a)>(b)?(a):(b))
    17 #define Min(a,b) ((a)<(b)?(a):(b))
    18 #define mem(a,b) memset(a,b,sizeof(a))
    19 #define lson l,mid,rt<<1
    20 #define rson mid+1,r,rt<<1|1
    21 const int N=70,INF=0x3f3f3f3f,MOD=1999997;
    22 const LL LLNF=0x3f3f3f3f3f3f3f3fLL;
    23 
    24 char s[12][N],ans[N];
    25 int next[N];
    26 int T,n,m,len;
    27 
    28 void getnext(char *a,int len)
    29 {
    30     int j=0,k=-1;
    31     next[0]=-1;
    32     while(j<len){
    33         if(k==-1 || a[k]==a[j])
    34             next[++j]=++k;
    35         else k=next[k];
    36     }
    37 }
    38 
    39 int cmp(char *a,int len)
    40 {
    41     int i,j,ok,k;
    42     for(i=1;i<n;i++){
    43         for(j=ok=k=0;j<m;j++){
    44             while(k>0 && s[i][j]!=a[k])k=next[k];
    45             if(s[i][j]==a[k])k++;
    46             if(k==len){ok=1;break;}
    47         }
    48         if(!ok)return 0;
    49     }
    50     return 1;
    51 }
    52 
    53 int main()
    54 {
    55  //   freopen("in.txt","r",stdin);
    56     int i,j,maxlen,ok,t;
    57     scanf("%d",&T);
    58     while(T--)
    59     {
    60         m=60;
    61         maxlen=-INF;
    62         scanf("%d",&n);
    63         for(i=0;i<n;i++)
    64             scanf("%s",s[i]);
    65         for(i=0;i<m;i++){
    66             for(j=i+3;j<=m;j++){
    67                 len=j-i;
    68                 if(len<maxlen)continue;
    69                 getnext(*s+i,len);
    70                 if(cmp(*s+i,len)){
    71                     t=s[0][j];
    72                     s[0][j]='\0';
    73                     if(len>maxlen){
    74                         maxlen=len;
    75                         strcpy(ans,*s+i);
    76                     }
    77                     else if(strcmp(ans,*s+i)>0)
    78                         strcpy(ans,*s+i);
    79                     s[0][j]=t;
    80                 }
    81             }
    82         }
    83 
    84         printf("%s\n",maxlen>2?ans:"no significant commonalities");
    85     }
    86     return 0;
    87 }
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  • 原文地址:https://www.cnblogs.com/zhsl/p/2837830.html
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