BZOJ 4566: [Haoi2016]找相同字符

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多串SAM

两串中相同的字串会在同一节点

每个节点搞个siz表示这个节点代表的字串出现了多少次

当然还要加一维，表示在哪个串中出现的次数

于是答案就是sigma (len[i]-len[par[i]])*siz[0][i]*siz[1][i]了

#include<cstdio>#include<iostream>#include<cstring>#include<cmath>#include<queue>#include<vector>#include<algorithm>#include<map>#include<set>#include<stack>#define rep(i,l,r) for(int i=l;i<=r;i++)#define per(i,r,l) for(int i=r;i>=l;i--)#define mmt(a,v) memset(a,v,sizeof(a))#define tra(i,u) for(int i=head[u];i;i=e[i].next)using namespace std;typedef long long ll;const int N=200000+5;const int M=800000+5;int par[M],ch[M][26],siz[2][M],len[M],root,sz,last;void init(){last=root=sz=1;}void extend(int x){int p=last,np=++sz;len[np]=len[p]+1;for(;p&&!ch[p][x];p=par[p])ch[p][x]=np;if(!p)par[np]=root;else{int q=ch[p][x];if(len[q]==len[p]+1)par[np]=q;else{int nq=++sz;len[nq]=len[p]+1;memcpy(ch[nq],ch[q],sizeof(ch[q]));par[nq]=par[q];par[q]=par[np]=nq;for(;p&&ch[p][x]==q;p=par[p])ch[p][x]=nq;}}last=np;}int deg[M];void toposort(){rep(i,1,sz)deg[par[i]]++;queue<int>q;rep(i,1,sz)if(!deg[i])q.push(i);while(!q.empty()){int u=q.front();q.pop();siz[0][par[u]]+=siz[0][u];siz[1][par[u]]+=siz[1][u];deg[par[u]]--;if(!deg[par[u]])q.push(par[u]);}}char s[N];int main(){//freopen("a.in","r",stdin);scanf("%s",s+1);int n=strlen(s+1);init();int p=root;rep(i,1,n){extend(s[i]-'a');p=ch[p][s[i]-'a'];siz[0][p]++;}scanf("%s",s+1);n=strlen(s+1);last=root;p=root;rep(i,1,n){extend(s[i]-'a');p=ch[p][s[i]-'a'];siz[1][p]++;}toposort();ll ans=0;rep(i,1,sz)if(par[i])ans+=1LL*(len[i]-len[par[i]])*siz[0][i]*siz[1][i];printf("%lld\n",ans);return 0;}

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