字符串 hash + poj 2774 (hash+二分）

  这题是求最长公共子串的长度，用二分求，check mid的时候，把s1的全部长度为mid的子串的hash值存入ihash容器，然后再求依次求出S2的长度为mid的子串的hash值，再用二分查找在ihash里找，找到了说明存在长度为mid的公共子串，找不到就存在。

这题存s1的hash值的时候用数组存就wa，就vector就ac，很奇怪

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<vector>
#include<algorithm>
using namespace std;
#define ull unsigned long long
const int maxn=100000+10;
ull ba[maxn],base=131;
vector<ull>ihash;
string s1,s2;
int len1,len2,ans;

bool check(int len)
{
    ihash.clear();
    ull tmp=0;
    for(int i=1;i<=len;i++)//求出s1里全部长度为len的子串的hash值
    {
        tmp=tmp*base+s1[i];
    }
    ihash.push_back(tmp);
    for(int i=2;i<=len1-len+1;i++)//求出s1里全部长度为len的子串的hash值
    {
        tmp=(tmp-s1[i-1]*ba[len-1])*base+s1[i+len-1];
        ihash.push_back(tmp);
    }
    sort(ihash.begin(),ihash.end());//为后面二分查找做准备
    //cout<<"s1 ihash："<<endl;
    //for(int i=0;i<tot;i++)
        //cout<<ihash[i]<<" ";
    tmp=0;
    for(int i=1;i<=len;i++)//求出s2里全部长度为len的子串的hash值
    {
        tmp=tmp*base+s2[i];
    }
    //cout<<"s2 ihash:"<<endl;
    //cout<<tmp<<" ";
    if(binary_search(ihash.begin(),ihash.end(),tmp))
    {
        //cout<<"find:"<<tmp<<endl;
        return true;
    }
    for(int i=2;i<=len2-len+1;i++)
    {
        tmp=(tmp-s2[i-1]*ba[len-1])*base+s2[i+len-1];
        //cout<<tmp<<" ";
        if(binary_search(ihash.begin(),ihash.end(),tmp))
        {
           // cout<<"find:"<<tmp<<endl;
            return true;
        }


    }
    return false;


}
int main()
{
    cin>>s1>>s2;
    len1=s1.size();
    len2=s2.size();
    //cout<<s1<<endl<<s2<<endl<<len1<<" "<<len2<<endl;
    s1=" "+s1;
    s2=" "+s2;
    int low=0,up=min(len1,len2),mid;
    ba[0]=1;
    for(int i=1;i<=up;i++)
        ba[i]=ba[i-1]*base;

    ans=-1;
    while(low<=up)
    {
        mid=(low+up)>>1;
        //cout<<"mid:"<<mid<<"  low:"<<low<<"  up:"<<up<<endl;
        if(check(mid))
        {

            ans=mid;
            low=mid+1;
        }
        else
            up=mid-1;
        //system("pause");
    }
    cout<<ans<<endl;
    return 0;
}