hdu 1403 Longest Common Substring (后缀数组)

Longest Common Substring

Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5776    Accepted Submission(s): 2052

Problem Description

Given two strings, you have to tell the length of the Longest Common Substring of them.

For example:
str1 = banana
str2 = cianaic

So the Longest Common Substring is "ana", and the length is 3.

 

Input

The input contains several test cases. Each test case contains two strings, each string will have at most 100000 characters. All the characters are in lower-case.

Process to the end of file.

 

Output

For each test case, you have to tell the length of the Longest Common Substring of them.

 

Sample Input

bananacianaic

 

Sample Output

3

 

题目的意思是求两个字符串的最大匹配长度

思路  将两个串A .B用一个绝对小的字符连接成一个串C

        求串C的后缀数组  在在这个里面找出两个分别代表A B 串的公共前缀的最长的那个

         (数据蛮水 这样居然过了)

#include<iostream>#include<algorithm>#include<cstdlib>#include<cctype>#include<cstdio>#include<string>#include<cstring>#include<vector>#include<set>#include<map>#include<queue>#include<cmath>#include<bitset>#define pi acos(-1.0)#define inf 1<<29#define INF 0x3fffffff#define zero 1e-8const int li[]={-1,0,1,0};const int lj[]={0,-1,0,1};using std::cout;using std::endl;const int N=200000+107;char str[2*N];int len1,len2,len;int rank[2 * N], SA[2 * N];int tem[2 * N];int ccnt[N];int height[N];inline bool cmp(int i,int cnt){    return rank[SA[i]+cnt]==rank[SA[i-1]+cnt]&&rank[SA[i]]==rank[SA[i-1]];}inline bool judge(int i,int j){     return (i<len1&&j>len1)||(i>len1&&j<len1);}void show(){    str[len]='\0';    for (int i=0 ;i<=len;++i)        std::cout<<i<<" SA: "<<SA[i]<<" "<<str+SA[i]<<std::endl;//<<" rank: "<<rank[i]<<std::endl;}void da(int MaxDataOfBegin){    memset(tem,0,sizeof(tem));    for (int i = 0; i < len; ++i){        tem[str[i]] = true;    }    for (int i = 1; i < MaxDataOfBegin+1; ++i){        tem[i] += tem[i - 1];    }    for (int i = 0; i < len; ++i)        rank[i] = tem[str[i]];    for (int i = len; i < 2 * len; ++i)        rank[i] = 0;    for (int lon = 2, cnt = 1; ; lon <<= 1, cnt<<=1) {        for (int i=0;i<=len;ccnt[i++]=0);        for (int i = cnt; i < len + cnt; ++i)            ccnt[rank[i]]++;        for (int i = 1; i <= len; ++i){            ccnt[i] += ccnt[i - 1];        }        for (int i = len + cnt-1; i >= cnt; --i) {            tem[ccnt[rank[i]]-1] = i - cnt;            ccnt[rank[i]]--;        }        for (int i=0;i<=len;ccnt[i++]=0);        for (int i = 0; i < len; ++i)            ccnt[rank[tem[i]]]++;        for (int i = 1; i <= len; ++i)            ccnt[i] += ccnt[i - 1];        for (int i = len-1; i >=0; --i) {            SA[ccnt[rank[tem[i]]]-1] = tem[i];            ccnt[rank[tem[i]]]--;        }        int rk=1;        tem[SA[0]] = rk;        for (int i=1;i<len;++i){            if (!cmp(i,cnt)) rk++;            tem[SA[i]]=rk;        }        for (int i=0;i<len;++i)            rank[i]=tem[i];        if (lon>=len) break;    }}int calheight(){    for (int i=0;i<=len;height[i++]=0);    for (int i=2;i<len;++i){        if (judge(SA[i-1],SA[i])){            int lon=0;            for (;(SA[i]+lon<len1||SA[i-1]+lon<len1)                    &&str[SA[i]+lon]==str[SA[i-1]+lon];lon++);                 height[i]=height[i]>lon?height[i]:lon;                 height[i-1]=lon;        }    }    int maxx=0;    for(int i=1;i<len;++i)        maxx=maxx>height[i]?maxx:height[i];    return maxx;}int main(){  while(~scanf("%s",str)){     len1=strlen(str);     str[len1]='Q';     scanf("%s",str+len1+1);     len=strlen(str);     da('z');     printf("%d\n",calheight());  }  return 0;}