Long Long Message

Problem's Link：

Mean:

求两个字符串的最长公共子串的长度。

analyse:

前面在学习后缀数组的时候已经做过一遍了，但是现在主攻字符串hash，再用字符串hash写一遍。

这题的思路是这样的：

1)取较短的串的长度作为high，然后二分答案(每次判断长度为mid=(low+high)>>1是否存在，如果存在就增加下界；不存在就缩小上界)；

2)主要是对答案的判断(judge函数)。具体参看代码注释。

Time complexity：O(n)

Source code：

// Memory   Time// 1347base     0MS// by : Snarl_jsb// 2014-10-04-21.16#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            #include
            
             #include
             #include
              
               #include
               
                #include
                
                 #define ULL unsigned long longusing namespace std;string s1,s2;int l1,l2,seed=131;vector
                 
                   hash;bool judge(int x){ hash.clear(); ULL tmp=0; for (int i = 0; i < x; i++) { tmp=tmp* seed + s1[i]; } hash.push_back(tmp); ULL base =1; for (int i = 1; i < x; i++) { base *= seed; } for (int i = x; i < l1; i++) { tmp=(tmp*seed+s1[i])-base*s1[i-x]*seed; hash.push_back(tmp); } sort(hash.begin(),hash.end()); ULL hashval = 0; for (int i = 0; i < x; i++) { hashval = hashval * seed + s2[i]; } if (binary_search(hash.begin(),hash.end(),hashval)) return 1; for (int i = x; i < l2; i++) { hashval = (hashval-(s2[i-x])*base)*seed+s2[i]; if (binary_search(hash.begin(),hash.end(),hashval)) return 1; } return 0;}int main(){ while (cin>>s1>>s2) { l1=s1.size(); l2=s2.size(); int ans = 0; int high = min(l1,l2); int low = 0; while (low <= high) { int mid = (low+high)>>1; if (judge(mid)) { ans = mid; low = mid+1; } else high = mid-1; } printf("%d\n",ans); } return 0;}

注释代码：

// Memory   Time// 1347k     0MS// by : Snarl_jsb// 2014-10-04-21.16#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            #include
            
             #include
             #include
              
               #include
               
                #include
                
                 #define ULL unsigned long longusing namespace std;string s1,s2;int l1,l2,seed=131;vector
                 
                   hash;bool judge(int x){ hash.clear();//多组数据时不要忘了清空全局数组 //构造s1串的hash表 ULL tmp=0; for (int i = 0; i < x; i++) { tmp=tmp* seed + s1[i]; } hash.push_back(tmp); ULL base =1; for (int i = 1; i < x; i++)//求出到达x的base值 { base *= seed; } for (int i = x; i < l1; i++) { tmp=(tmp*seed+s1[i])-base*s1[i-x]*seed; hash.push_back(tmp); } //构造完毕 sort(hash.begin(),hash.end()); //二分查找加速，必需先排序 ULL hashval = 0; for (int i = 0; i < x; i++)//求出s2串0到x的hash值 { hashval = hashval * seed + s2[i]; } if (binary_search(hash.begin(),hash.end(),hashval))//查找s2串0到x的hash值是否在s1串的hash表中 return 1; for (int i = x; i < l2; i++)//如果上面的s2串0到x的hash值未匹配成功，这儿接着匹配s2串长度为x的hash值是否出现在s1串的hash表中 { hashval = hashval*seed+s2[i]-s2[i-x]*base*seed; if (binary_search(hash.begin(),hash.end(),hashval)) return 1; } return 0;}int main(){ while (cin>>s1>>s2) { l1=s1.size(); l2=s2.size(); int ans = 0; int low=0,high = min(l1,l2); while (low <= high)//二分答案 { int mid = (low+high)>>1; if (judge(mid))//判断答案是否可行 { ans = mid; low = mid+1; } else high = mid-1; } printf("%d\n",ans); } return 0;}