Codeforces 163E && HDU 4117 && Noi2011阿狸的打字机 && boj 1602
http://codeforces.com/problemset/problem/163/E
http://acm.hdu.edu.cn/showproblem.php?pid=4117
http://61.187.179.132/JudgeOnline/problem.php?id=2434
http://n.boj.me/onlinejudge/newoj/showProblem/show_problem.php?problem_id=1602
这四道题均为用数据结构维护的AC自动机。
基本想法是建好AC自动机后，顺便建立fail树，然后用线段树维护操作。
基本原理：从某个节点沿着fail指针走直到根节点过程中遇到的节点都是改串的后缀相同的串，而fail指针构成了一棵fail树，所以一个串可能是哪些串的后缀字串呢？答案是节点对应fail树里的一颗子树。
Codeforces 163E
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#define N 1000100
using namespace std;
char S[N];//所以串读入S中
int str[N],mp[N],vis[N];//记录读入串的最后一个位置下标，mp[i]表示第i个单词的自动机编号，
int pos;
int next[N][26],fail[N],flag[N];
int newnode(){
memset(next[pos],0,sizeof(next[pos]));
fail[pos]=flag[pos]=0;
return pos++;
}
void insert(char * s,int id){
int p=0,i=0;
for(;s[i];i++){
int k=s[i]-'a';
p=next[p][k]?next[p][k]:next[p][k]=newnode();
}
flag[p]=1;
mp[id]=p;
}
void makefail(){
queue<int>q;
q.push(0);
while(!q.empty()){
int u=q.front();
q.pop();
for(int i=0;i<26;i++){
int v=next[u][i];
if(v==0) next[u][i]=next[fail[u]][i];
else q.push(v);
if(v && u)fail[v]=next[fail[u]][i];
}
}
}
struct Tree{
int l,r;
int add;
int sum;
}tree[N<<2];
void build(int s,int t,int id){
tree[id].l=s;
tree[id].r=t;
tree[id].add=tree[id].sum=0;
if(s!=t){
int mid=(s+t)>>1;
build(s,mid,id<<1);
build(mid+1,t,id<<1|1);
}
}
void update(int s,int t,int add,int id){
//printf("%d %d %d\n",s,t,id);
if(tree[id].l==s && tree[id].r==t){
tree[id].add+=add;
tree[id].sum+=add;
return ;
}
if(tree[id].add!=0){
tree[id<<1].add+=tree[id].add;
tree[id<<1|1].add+=tree[id].add;
tree[id<<1].sum+=tree[id].add;
tree[id<<1|1].sum+=tree[id].add;
tree[id].add=0;
}
int mid=(tree[id].l+tree[id].r)>>1;
if(t<=mid) update(s,t,add,id<<1);
else if(mid<s) update(s,t,add,id<<1|1);
else{
update(s,mid,add,id<<1);
update(mid+1,t,add,id<<1|1);
}
}
int query(int POS,int id){
if(tree[id].l==tree[id].r){
return tree[id].sum;
}
if(tree[id].add!=0){
tree[id<<1].add+=tree[id].add;
tree[id<<1|1].add+=tree[id].add;
tree[id<<1].sum+=tree[id].add;
tree[id<<1|1].sum+=tree[id].add;
tree[id].add=0;
}
int mid=(tree[id].l+tree[id].r)>>1;
if(POS<=mid) return query(POS,id<<1);
else return query(POS,id<<1|1);
}
struct Edge{
int v,next;
}edge[N*2];
int head[N],cnt;
int tim[N],son[N],dep;//tim[i]为自动机i个节点的dfs序
void init(){
memset(head,-1,sizeof(head));
cnt=dep=0;
}
void addedge(int u,int v){
edge[cnt].v=v;
edge[cnt].next=head[u];
head[u]=cnt++;
edge[cnt].v=u;
edge[cnt].next=head[v];
head[v]=cnt++;
}
void dfs(int u,int fa){
tim[u]=++dep;
son[u]=1;
for(int i=head[u];i!=-1;i=edge[i].next){
int v=edge[i].v;
if(v==fa)continue;
dfs(v,u);
son[u]+=son[v];
}
}
void build_failtree(){
for(int i=1;i<pos;i++) addedge(fail[i],i);
dfs(0,-1);
//for(int i=0;i<pos;i++) printf("[%d]=%d\n",i,tim[i]);
build(1,pos,1);
//update(1,pos-1,1,1);
for(int i=0;i<pos;i++) if(flag[i])
//printf("%d %d\n",tim[i],tim[i]+son[i]-1);
update(tim[i],tim[i]+son[i]-1,1,1);
}
int AC(char * s){
int ans=0,p=0;
for(int i=0;s[i];i++){
p=next[p][s[i]-'a'];
ans+=query(tim[p],1);
//printf("%d %d %d\n",p,tim[p],ans);
}
return ans;
}
int trans(char * s){
int ans=0;
for(int i=0;s[i];i++){
ans=ans*10+s[i]-'0';
}
return ans;
}
int main(){
int n,k;
pos=0;newnode();
str[0]=0;
scanf("%d %d",&n,&k);
for(int i=1;i<=k;i++){
scanf("%s",&S[str[i-1]+1]);
str[i]=str[i-1]+strlen(&S[str[i-1]+1]);
insert(&S[str[i-1]+1],i);
}
makefail();
init();
build_failtree();
for(int i=1;i<=n;i++) vis[i]=1;
//for(int i=0;i<pos;i++) printf("%d %d\n",i,query(tim[i],1));
//for(int i=0;i<pos;i++)printf("%d: %d %d %d\n",i,next[i][0],next[i][1],fail[i]);
for(int i=1;i<=n;i++){
scanf("%s",S);
if(S[0]=='+'){
int tem=trans(&S[1]);
if(!vis[tem]){
//printf("fuck\n");
update(tim[mp[tem]],tim[mp[tem]]+son[mp[tem]]-1,1,1);
vis[tem]=1;
}
}
else if(S[0]=='-'){
int tem=trans(&S[1]);
//printf("****%d\n",tem);
if(vis[tem]){
//printf("fuck %d %d\n",tim[tem],tim[tem]+son[tem]-1);
update(tim[mp[tem]],tim[mp[tem]]+son[mp[tem]]-1,-1,1);
vis[tem]=0;
}
}
else{
printf("%d\n",AC(&S[1]));
}
}
return 0;
}
HDU 4117
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#define N 300100
using namespace std;
char S[N];//所以串读入S中
int n,str[N],dp[N];//str[i]记录读入串的最后一个位置下标
int pos;
int next[N][26],fail[N];
int newnode(){
memset(next[pos],0,sizeof(next[pos]));
fail[pos]=0;
return pos++;
}
void insert(char * s,int id){
int p=0,i=0;
for(;s[i];i++){
int k=s[i]-'a';
p=next[p][k]?next[p][k]:next[p][k]=newnode();
}
}
void makefail(){
queue<int>q;
q.push(0);
while(!q.empty()){
int u=q.front();
q.pop();
for(int i=0;i<26;i++){
int v=next[u][i];
if(v==0) next[u][i]=next[fail[u]][i];
else q.push(v);
if(v && u)fail[v]=next[fail[u]][i];
}
}
}
struct Tree{
int l,r;
int lazy;
int MAX;
}tree[N<<2];
void build(int s,int t,int id){
tree[id].l=s;
tree[id].r=t;
tree[id].MAX=tree[id].lazy=0;
if(s!=t){
int mid=(s+t)>>1;
build(s,mid,id<<1);
build(mid+1,t,id<<1|1);
}
}
void update(int s,int t,int lazy,int id){
//printf("%d %d %d\n",s,t,id);
if(tree[id].l==s && tree[id].r==t){
tree[id].lazy=max(tree[id].lazy,lazy);
tree[id].MAX=max(tree[id].MAX,lazy);
return ;
}
if(tree[id].lazy!=0){
tree[id<<1].lazy=max(tree[id<<1].lazy,tree[id].lazy);
tree[id<<1|1].lazy=max(tree[id<<1|1].lazy,tree[id].lazy);
tree[id<<1].MAX=max(tree[id<<1].MAX,tree[id].lazy);
tree[id<<1|1].MAX=max(tree[id<<1|1].MAX,tree[id].lazy);
tree[id].lazy=0;
}
int mid=(tree[id].l+tree[id].r)>>1;
if(t<=mid) update(s,t,lazy,id<<1);
else if(mid<s) update(s,t,lazy,id<<1|1);
else{
update(s,mid,lazy,id<<1);
update(mid+1,t,lazy,id<<1|1);
}
}
int query(int POS,int id){
if(tree[id].l==tree[id].r){
return tree[id].MAX;
}
if(tree[id].lazy!=0){
tree[id<<1].lazy=max(tree[id<<1].lazy,tree[id].lazy);
tree[id<<1|1].lazy=max(tree[id<<1|1].lazy,tree[id].lazy);
tree[id<<1].MAX=max(tree[id<<1].MAX,tree[id].lazy);
tree[id<<1|1].MAX=max(tree[id<<1|1].MAX,tree[id].lazy);
tree[id].lazy=0;
}
int mid=(tree[id].l+tree[id].r)>>1;
if(POS<=mid) return query(POS,id<<1);
else return query(POS,id<<1|1);
}
struct Edge{
int v,next;
}edge[N*2];
int head[N],cnt;
int tim[N],son[N],dep;//tim[i]为自动机i个节点的dfs序
void init(){
memset(head,-1,sizeof(head));
cnt=dep=0;
}
void addedge(int u,int v){
edge[cnt].v=v;
edge[cnt].next=head[u];
head[u]=cnt++;
edge[cnt].v=u;
edge[cnt].next=head[v];
head[v]=cnt++;
}
void dfs(int u,int fa){
tim[u]=++dep;
son[u]=1;
for(int i=head[u];i!=-1;i=edge[i].next){
int v=edge[i].v;
if(v==fa)continue;
dfs(v,u);
son[u]+=son[v];
}
}
void build_failtree(){
for(int i=1;i<pos;i++) addedge(fail[i],i);
dfs(0,-1);
build(1,pos,1);
}
void solve(){
for(int i=1;i<=n;i++){
if(dp[i]>0){
int p=0,tem=0;
for(int j=str[i-1]+1;j<=str[i];j++){
p=next[p][S[j]-'a'];
tem=max(tem,query(tim[p],1));
}
dp[i]+=tem;
if(dp[i]>0) update(tim[p],tim[p]+son[p]-1,dp[i],1);
}
}
}
int main(){
int T;
scanf("%d",&T);
for(int t=1;t<=T;t++){
pos=0;newnode();
str[0]=0;
scanf("%d",&n);
for(int i=1;i<=n;i++){
scanf("%s %d",&S[str[i-1]+1],&dp[i]);
str[i]=str[i-1]+strlen(&S[str[i-1]+1]);
if(dp[i]>0)insert(&S[str[i-1]+1],i);
}
makefail();
init();
build_failtree();
solve();
int ans=0;
for(int i=1;i<=n;i++) ans=max(ans,dp[i]);
printf("Case #%d: %d\n",t,ans);
}
return 0;
}
Noi2011阿狸的打字机
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <algorithm>
#define N 100100
using namespace std;
char S[N];//所以串读入S中
vector<pair<int,int> >v[N];
int n,str[N],mp[N],ans[N];//str[i]记录读入串的最后一个位置下标，mp[i]记录第i个单词的自动机位置
int pos;
int next[N][26],fail[N],fa[N];//fa[i]表示i节点在insert时的父亲
int newnode(){
memset(next[pos],0,sizeof(next[pos]));
fail[pos]=fa[pos]=0;
return pos++;
}
void insert(char * s){
int p=0,i=0,cou=0;
for(;s[i];i++){
if(s[i]=='P'){
mp[++cou]=p;
}
else if(s[i]=='B'){
p=fa[p];
}
else{
int k=s[i]-'a',tem=p;
p=next[p][k]?next[p][k]:next[p][k]=newnode();
fa[p]=tem;
}
}
}
void makefail(){
queue<int>q;
q.push(0);
while(!q.empty()){
int u=q.front();
q.pop();
for(int i=0;i<26;i++){
int v=next[u][i];
if(v==0) next[u][i]=next[fail[u]][i];
else q.push(v);
if(v && u)fail[v]=next[fail[u]][i];
}
}
}
struct Tree{
int l,r;
int add;
int sum;
}tree[N<<2];
void build(int s,int t,int id){
tree[id].l=s;
tree[id].r=t;
tree[id].add=tree[id].sum=0;
if(s!=t){
int mid=(s+t)>>1;
build(s,mid,id<<1);
build(mid+1,t,id<<1|1);
}
}
void update(int POS,int add,int id){
tree[id].sum+=add;
if(tree[id].l==tree[id].r){
return ;
}
if(tree[id].add!=0){
tree[id<<1].add+=tree[id].add;
tree[id<<1|1].add+=tree[id].add;
tree[id<<1].sum+=tree[id].add;
tree[id<<1|1].sum+=tree[id].add;
tree[id].add=0;
}
int mid=(tree[id].l+tree[id].r)>>1;
if(POS<=mid) update(POS,add,id<<1);
else update(POS,add,id<<1|1);
}
int query(int s,int t,int id){
if(tree[id].l==s && tree[id].r==t){
return tree[id].sum;
}
if(tree[id].add!=0){
tree[id<<1].add+=tree[id].add;
tree[id<<1|1].add+=tree[id].add;
tree[id<<1].sum+=tree[id].add;
tree[id<<1|1].sum+=tree[id].add;
tree[id].add=0;
}
int mid=(tree[id].l+tree[id].r)>>1;
if(t<=mid) return query(s,t,id<<1);
else if(s>mid) return query(s,t,id<<1|1);
else return query(s,mid,id<<1)+query(mid+1,t,id<<1|1);
}
struct Edge{
int v,next;
}edge[N*2];
int head[N],cnt;
int tim[N],son[N],dep;//tim[i]为自动机i个节点的dfs序
void init(){
memset(head,-1,sizeof(head));
cnt=dep=0;
}
void addedge(int u,int v){
edge[cnt].v=v;
edge[cnt].next=head[u];
head[u]=cnt++;
edge[cnt].v=u;
edge[cnt].next=head[v];
head[v]=cnt++;
}
void dfs(int u,int fa){
tim[u]=++dep;
son[u]=1;
for(int i=head[u];i!=-1;i=edge[i].next){
int v=edge[i].v;
if(v==fa)continue;
dfs(v,u);
son[u]+=son[v];
}
}
void build_failtree(){
for(int i=1;i<pos;i++) addedge(fail[i],i);
dfs(0,-1);
build(1,pos,1);
}
void solve(char * s){
int p=0,cou=0;
for(int i=0;s[i];i++){
if(s[i]=='P'){
++cou;
for(int j=0;j<v[cou].size();j++){
int x=v[cou][j].first;
int y=v[cou][j].second;
//printf("%d %d~%d\n",y,tim[mp[x]],tim[mp[x]]+son[mp[x]]-1);
ans[y]=query(tim[mp[x]],tim[mp[x]]+son[mp[x]]-1,1);
}
}
else if(s[i]=='B'){
//printf("%d~%d -1\n",tim[p],tim[p]);
update(tim[p],-1,1);
p=fa[p];
}
else{
p=next[p][s[i]-'a'];
//printf("%d~%d +1\n",tim[p],tim[p]);
update(tim[p],1,1);
}
}
}
int main(){
int m,x,y;
scanf("%s",S);
scanf("%d",&m);
for(int i=0;i<N;i++) v[i].clear();
for(int i=1;i<=m;i++){
scanf("%d %d",&x,&y);
v[y].push_back(make_pair(x,i));
}
pos=0;newnode();
insert(S);
makefail();
init();
build_failtree();
//for(int i=0;i<pos;i++) printf("**%d %d\n",i,tim[i]);
solve(S);
for(int i=1;i<=m;i++) printf("%d\n",ans[i]);
return 0;
}
BOJ 1602这个题是2013年四川省赛题目
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#define N 1000100
using namespace std;
char S[N*2];//所以串读入S中
int n,str[N],type[N],mp[N];//记录读入串的最后一个位置下标，mp[i]表示第i个单词的自动机编号，
int pos;
int next[N][26],fail[N];
int newnode(){
memset(next[pos],0,sizeof(next[pos]));
fail[pos]=0;
return pos++;
}
void insert(char * s,int id){
int p=0,i=0;
for(;s[i];i++){
int k=s[i]-'a';
p=next[p][k]?next[p][k]:next[p][k]=newnode();
}
mp[id]=p;
}
void makefail(){
queue<int>q;
q.push(0);
while(!q.empty()){
int u=q.front();
q.pop();
for(int i=0;i<26;i++){
int v=next[u][i];
if(v==0) next[u][i]=next[fail[u]][i];
else q.push(v);
if(v && u)fail[v]=next[fail[u]][i];
}
}
}
struct Tree{
int l,r;
int add;
int sum;
}tree[N<<2];
void build(int s,int t,int id){
tree[id].l=s;
tree[id].r=t;
tree[id].add=tree[id].sum=0;
if(s!=t){
int mid=(s+t)>>1;
build(s,mid,id<<1);
build(mid+1,t,id<<1|1);
}
}
void update(int s,int t,int add,int id){
if(tree[id].l==s && tree[id].r==t){
tree[id].add+=add;
tree[id].sum+=add;
return ;
}
if(tree[id].add!=0){
tree[id<<1].add+=tree[id].add;
tree[id<<1|1].add+=tree[id].add;
tree[id<<1].sum+=tree[id].add;
tree[id<<1|1].sum+=tree[id].add;
tree[id].add=0;
}
int mid=(tree[id].l+tree[id].r)>>1;
if(t<=mid) update(s,t,add,id<<1);
else if(mid<s) update(s,t,add,id<<1|1);
else{
update(s,mid,add,id<<1);
update(mid+1,t,add,id<<1|1);
}
}
int query(int POS,int id){
if(tree[id].l==tree[id].r){
return tree[id].sum;
}
if(tree[id].add!=0){
tree[id<<1].add+=tree[id].add;
tree[id<<1|1].add+=tree[id].add;
tree[id<<1].sum+=tree[id].add;
tree[id<<1|1].sum+=tree[id].add;
tree[id].add=0;
}
int mid=(tree[id].l+tree[id].r)>>1;
if(POS<=mid) return query(POS,id<<1);
else return query(POS,id<<1|1);
}
struct Edge{
int v,next;
}edge[N*2];
int head[N],cnt;
int tim[N],son[N],dep;//tim[i]为自动机i个节点的dfs序
void init(){
memset(head,-1,sizeof(head));
cnt=dep=0;
}
void addedge(int u,int v){
edge[cnt].v=v;
edge[cnt].next=head[u];
head[u]=cnt++;
edge[cnt].v=u;
edge[cnt].next=head[v];
head[v]=cnt++;
}
void dfs(int u,int fa){
tim[u]=++dep;
son[u]=1;
for(int i=head[u];i!=-1;i=edge[i].next){
int v=edge[i].v;
if(v==fa)continue;
dfs(v,u);
son[u]+=son[v];
}
}
void build_failtree(){
for(int i=1;i<pos;i++) addedge(fail[i],i);
dfs(0,-1);
build(1,pos,1);
}
void solve(){
for(int i=1;i<=n;i++){
if(type[i]==1){
update(tim[mp[i]],tim[mp[i]]+son[mp[i]]-1,1,1);
}
else{
int p=0;
long long ans=0;
for(int j=str[i-1]+1;j<=str[i];j++){
p=next[p][S[j]-'a'];
ans+=(long long)query(tim[p],1);
}
printf("%lld\n",ans);
}
}
}
int main(){
int t,T;
char ss[10];
scanf("%d",&T);
for(t=1;t<=T;t++){
pos=0;newnode();
str[0]=0;
scanf("%d %d",&n);
memset(mp,0,sizeof(mp));
for(int i=1;i<=n;i++){
scanf("%s %s",ss,&S[str[i-1]+1]);
str[i]=str[i-1]+strlen(&S[str[i-1]+1]);
if(!strcmp(ss,"ask")) type[i]=2;
else type[i]=1;
if(type[i]==1) insert(&S[str[i-1]+1],i);
}
makefail();
init();
build_failtree();
solve();
}
return 0;
}