字符串双哈希+自定义unordered_map pair<int,int>当key 1923. 最长公共子路径

class Solution {
public:
    typedef pair<int,int> pii;
    const int k1=1331;
    const int k2=13331;
    const int mod1=1e9+7;
    const int mod2=1e9+9;
    int p1[100005],p2[100005];
     struct pairhash {
        size_t operator() (const pair<int, int>& p) const {
            auto fn = hash<int>();
            return (fn(p.first) <<16) ^ fn(p.second);
        }
    };

    int longestCommonSubpath(int n, vector<vector<int>>& paths) {
         p1[0]=p2[0]=1;
        for(int i=1;i<=100001;i++)
        {
            p1[i]=1ll*p1[i-1]*k1%mod1;
            p2[i]=1ll*p2[i-1]*k2%mod2;
        }
        int row=paths.size();
        vector<int>s1[row+1];
        vector<int>s2[row+1];
        for(int i=1;i<=row;i++)s2[i]=s1[i]=vector<int>(paths[i-1].size()+1);
        for(int i=1;i<=row;i++)
        {
            int sz=paths[i-1].size();
            for(int j=1;j<=sz;j++)
            {
                s1[i][j]=((1ll*s1[i][j-1]*k1%mod1)+paths[i-1][j-1])%mod1;
                s2[i][j]=((1ll*s2[i][j-1]*k2%mod2)+paths[i-1][j-1])%mod2;
            }
        }

        function<int(int)>check=[&](int len)
        {
            unordered_map<pii,int,pairhash>mp;
           
            for(int i=1;i+len-1<=paths[0].size();i++)
            {
                 int j=i+len-1;
                 int x1=((s1[1][j]-(1ll*s1[1][i-1]*p1[j-i+1]%mod1))%mod1+mod1)%mod1;
                 int x2=((s2[1][j]-(1ll*s2[1][i-1]*p2[j-i+1]%mod2))%mod2+mod2)%mod2;
                 mp[{x1,x2}]=1;
            }

            for(int x=2;x<=row;x++)
            {
                unordered_map<pii,int,pairhash>vis;
                int sz=paths[x-1].size();
                bool ok=false;
                for(int i=1;i+len-1<=sz;i++)
                {
                    int j=i+len-1;
                    int t1=((s1[x][j]-(1ll*s1[x][i-1]*p1[j-i+1]%mod1))%mod1+mod1)%mod1;
                    int t2=((s2[x][j]-(1ll*s2[x][i-1]*p2[j-i+1]%mod2))%mod2+mod2)%mod2;
                    if(mp[{t1,t2}]&&!vis[{t1,t2}]){mp[{t1,t2}]++;vis[{t1,t2}]=1;}
                }    
            }
            for(auto &p:mp)if(p.second==row)return 1;
            return 0;
        };
        int l=0,r=1e6;
        while(l<=r)
        { 
            int mid=(l+r)>>1;
            if(check(mid))l=mid+1;
            else r=mid-1;
        }
        return r;
    }
};