Long Long Message
Problem's Link：http://poj.org/problem?id=2774
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<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<string>
#include<climits>
#include<cmath>
#define ULL unsigned long long
using namespace std;
string s1,s2;
int l1,l2,seed=131;
vector<ULL> 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<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<string>
#include<climits>
#include<cmath>
#define ULL unsigned long long
using namespace std;
string s1,s2;
int l1,l2,seed=131;
vector<ULL> 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;
}
