标签：数据 nta sap mis for 保存 uil The ref
InputThe first line of the input gives the number of test cases T; T test cases follow.
Each test case contains two integer N and Q, The number of words in the dictionary, and the number of words in the list.
Next N line, each line has a string Wi, represents the ith word in the dictionary (0<|Wi|≤100000 0<|Wi|≤100000
)
Next Q line, each line has two string Pi , Si, represents the prefix and suffix of the ith word in the list (0<|Pi|,|Si|≤100000,0<|Pi|+|Si|≤100000 0<|Pi|,|Si|≤100000,0<|Pi|+|Si|≤100000
)
All of the above characters are lowercase letters.
The dictionary does not contain the same words.
Limits
T≤5 T≤5
0<N,Q≤100000 0<N,Q≤100000
∑Si+Pi≤500000 ∑Si+Pi≤500000
∑Wi≤500000 ∑Wi≤500000
OutputFor each test case, output Q lines, an integer per line, represents the answer to each word in the list.
Sample Input
1 4 4 aba cde acdefa cdef a a cd ef ac a ce f
Sample Output
2 1 1 0
题意：已知N个单词，Q次询问，每次询问给出pre和suf，统计有多少个单词的前缀为pre，后缀为suf，而且要满足二者不相交。
思路：我们把询问建立AC自动机，单词用来跑AC自动机，跑到了就累计。
合理建立AC自动机的方式为：每个询问转为为 suf+‘{‘+pre；
跑AC自动机的方式为： 每个单词转化为 S+’{‘+S；
跑的时候如果fail可以走到某个询问，说明这个询问是这里的前后缀。（AC了但是不严谨的代码）
#include<bits/stdc++.h> #define rep(i,a,b) for(int i=a;i<=b;i++) using namespace std; const int maxn=500010; char c[maxn],s[maxn],pre[maxn],suf[maxn]; int tot,F[maxn],L[maxn],ch[maxn][27],cnt,pos[maxn]; int N,Q,dep[maxn],sum[maxn],fail[maxn],q[maxn],head,tail; void insert(int opt){ int Now=0,len1=strlen(suf+1),len2=strlen(pre+1); rep(i,1,len1){ if(!ch[Now][suf[i]-‘a‘]) ch[Now][suf[i]-‘a‘]=++cnt,sum[cnt]=0; Now=ch[Now][suf[i]-‘a‘]; } if(!ch[Now][26]) ch[Now][26]=++cnt,sum[cnt]=0; Now=ch[Now][26]; rep(i,1,len2){ if(!ch[Now][pre[i]-‘a‘]) ch[Now][pre[i]-‘a‘]=++cnt,sum[cnt]=0; Now=ch[Now][pre[i]-‘a‘]; } pos[opt]=Now; dep[Now]=len1+len2; } void buildfail() { head=tail=0; for(int i=0;i<=26;i++) if(ch[0][i]) q[++head]=ch[0][i]; while(tail<head){ int Now=q[++tail]; for(int i=0;i<=26;i++){ if(ch[Now][i]) { fail[ch[Now][i]]=ch[fail[Now]][i]; q[++head]=ch[Now][i]; } else ch[Now][i]=ch[fail[Now]][i]; } } } void solve(int B,int len) { int Now=0; rep(i,B+1,B+len) Now=ch[Now][c[i]-‘a‘]; Now=ch[Now][26]; rep(i,B+1,B+len){ Now=ch[Now][c[i]-‘a‘]; int tmp=Now; while(dep[tmp]>len) tmp=fail[tmp]; sum[tmp]++; } } int main() { int T; scanf("%d",&T); while(T--){ tot=cnt=0; memset(fail,0,sizeof(fail)); memset(ch,0,sizeof(ch)); scanf("%d%d",&N,&Q); rep(i,1,N){ scanf("%s",s+1); L[i]=strlen(s+1); F[i]=tot; rep(j,1,L[i]) c[++tot]=s[j]; //保存单词 } rep(i,1,Q){ scanf("%s%s",pre+1,suf+1); insert(i); } buildfail(); rep(i,1,N) solve(F[i],L[i]); for(int i=cnt;i>=1;i--) sum[fail[q[i]]]+=sum[q[i]]; //累加前缀和 rep(i,1,Q) printf("%d\n",sum[pos[i]]); } return 0; }
虽然上面的代码AC了，但是我感觉是可以hack掉，应该是数据比较水。 因为一个单词对一个询问最多有一个贡献，而这样跑下来有的单词的贡献可能大于1，所以我们加一个时间戳，保证每个单词的贡献最多为1。
#include<bits/stdc++.h> #define rep(i,a,b) for(int i=a;i<=b;i++) using namespace std; const int maxn=500010; char c[maxn],s[maxn],pre[maxn],suf[maxn]; int tot,F[maxn],L[maxn],ch[maxn][27],cnt,pos[maxn],Laxt[maxn]; int N,Q,dep[maxn],sum[maxn],fail[maxn],q[maxn],head,tail; void insert(int opt){ int Now=0,len1=strlen(suf+1),len2=strlen(pre+1); rep(i,1,len1){ if(!ch[Now][suf[i]-‘a‘]) ch[Now][suf[i]-‘a‘]=++cnt,sum[cnt]=0; Now=ch[Now][suf[i]-‘a‘]; } if(!ch[Now][26]) ch[Now][26]=++cnt,sum[cnt]=0; Now=ch[Now][26]; rep(i,1,len2){ if(!ch[Now][pre[i]-‘a‘]) ch[Now][pre[i]-‘a‘]=++cnt,sum[cnt]=0; Now=ch[Now][pre[i]-‘a‘]; } pos[opt]=Now; dep[Now]=len1+len2; } void buildfail() { head=tail=0; for(int i=0;i<=26;i++) if(ch[0][i]) q[++head]=ch[0][i]; while(tail<head){ int Now=q[++tail]; for(int i=0;i<=26;i++){ if(ch[Now][i]) { fail[ch[Now][i]]=ch[fail[Now]][i]; q[++head]=ch[Now][i]; } else ch[Now][i]=ch[fail[Now]][i]; } } } void solve(int time,int B,int len) { int Now=0; rep(i,B+1,B+len) Now=ch[Now][c[i]-‘a‘]; Now=ch[Now][26]; rep(i,B+1,B+len){ Now=ch[Now][c[i]-‘a‘]; int tmp=Now; while(tmp) { if(Laxt[tmp]==time) break; Laxt[tmp]=time;//加一个时间戳，保证每个单词的贡献最多为1 if(dep[tmp]<=len) sum[tmp]++; tmp=fail[tmp]; } } } int main() { int T; scanf("%d",&T); while(T--){ tot=cnt=0; memset(fail,0,sizeof(fail)); memset(ch,0,sizeof(ch)); memset(Laxt,0,sizeof(Laxt)); scanf("%d%d",&N,&Q); rep(i,1,N){ scanf("%s",s+1); L[i]=strlen(s+1); F[i]=tot; rep(j,1,L[i]) c[++tot]=s[j]; //保存单词 } rep(i,1,Q){ scanf("%s%s",pre+1,suf+1); insert(i); } buildfail(); rep(i,1,N) solve(i,F[i],L[i]); rep(i,1,Q) printf("%d\n",sum[pos[i]]); } return 0; }
HDU - 6096 ：String （AC自动机，已知前后缀，匹配单词，弱数据）
标签：数据 nta sap mis for 保存 uil The ref
原文地址：https://www.cnblogs.com/hua-dong/p/9807370.html
