NOI2019考前做NOI2018题。。
题目链接: (bzoj) https://www.lydsy.com/JudgeOnline/problem.php?id=5417
(luogu) https://www.luogu.org/problemnew/show/P4770
(uoj) http://uoj.ac/problem/395
题解: 其实非常水，转化成\(S[l,r]\)和\(T\)有多少本质不同的公共子串，我们就是要求出串\(T\)每个位置与\(S[l,r]\)最长匹配的长度。那么就对\(S\)建SAM，把\(T\)在\(S\)上跑，每次到达一个点之后如果它的子树内没有\([l,r]\)之间的点就跳fa. 求子树内的点(即Right集合)可以自上而下线段树合并。因为要求本质不同的公共子串个数，那么对\(T\)也建立后缀自动机，统计T的后缀自动机上每个点与S的公共子串个数。细节挺多，挂了好几次，详见代码。
时间复杂度\(O(n\log n)\), \(n\)为输入字符串总长
字符串细节真的好多啊，提醒自己明天如果写字符串题一定要对拍！(快算了吧，我必不可能会做字符串)
代码
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cassert>
#include<iostream>
#include<algorithm>
#define llong long long
using namespace std;inline int read()
{int x=0; bool f=1; char c=getchar();for(;!isdigit(c);c=getchar()) if(c=='-') f=0;for(; isdigit(c);c=getchar()) x=(x<<3)+(x<<1)+(c^'0');if(f) return x;return -x;
}const int N = 5e5;
const int S = 26;
const int LGN = 21;
struct SuffixAutomaton
{int fa[(N<<1)+3];int len[(N<<1)+3];int son[(N<<1)+3][S+1];int ord[(N<<1)+3];int buc[(N<<1)+3];int rtn,siz,lstpos;void init(){for(int i=1; i<=siz; i++){len[i] = fa[i] = 0;for(int j=1; j<=S; j++) son[i][j] = 0;}rtn = siz = lstpos = 1;}int insertchar(int ch){int p = lstpos,np; siz++; np = lstpos = siz; len[np] = len[p]+1;for(; p && son[p][ch]==0; p=fa[p]) {son[p][ch] = np;}if(p==0) {fa[np] = 1;}else{int q = son[p][ch];if(len[q]==len[p]+1) {fa[np] = q;}else{siz++; int nq = siz; len[nq] = len[p]+1;memcpy(son[nq],son[q],sizeof(son[q]));fa[nq] = fa[q]; fa[np] = fa[q] = nq;for(; p && son[p][ch]==q; p=fa[p]) {son[p][ch] = nq;}}}return np;}void getord(){for(int i=1; i<=siz; i++) buc[i] = 0;for(int i=1; i<=siz; i++) buc[len[i]]++;for(int i=1; i<=siz; i++) buc[i] += buc[i-1];for(int i=siz; i>=1; i--) ord[buc[len[i]]--] = i;}
} sam,samt;
struct SegmentTree
{int ls,rs;
} sgt[(N<<1)*LGN+3];
char a[N+3],b[N+3];
int mxl[(N<<1)+3];
int rt[(N<<1)+3];
int n,m,q,rtn,siz;void addval(int &pos,int le,int ri,int lrb)
{siz++; sgt[siz] = sgt[pos]; pos = siz;if(le==lrb && ri==lrb) {return;}int mid = (le+ri)>>1;if(lrb<=mid) {addval(sgt[pos].ls,le,mid,lrb);}else {addval(sgt[pos].rs,mid+1,ri,lrb);}
}int merge(int pos1,int pos2)
{if(pos1==0) return pos2; if(pos2==0) return pos1;siz++; int pos = siz;sgt[pos].ls = merge(sgt[pos1].ls,sgt[pos2].ls);sgt[pos].rs = merge(sgt[pos1].rs,sgt[pos2].rs);return pos;
}int query(int pos,int le,int ri,int rb) //largest <=rb
{if(pos==0) return 0;if(le==ri) return le;int mid = (le+ri)>>1;if(rb<=mid) return query(sgt[pos].ls,le,mid,rb);else{int tmp = query(sgt[pos].rs,mid+1,ri,rb);return tmp ? tmp : query(sgt[pos].ls,le,mid,rb);}
}int main()
{scanf("%s",a+1); n = strlen(a+1);for(int i=1; i<=n; i++) a[i] -= 96;sam.init();rtn = siz = 1;for(int i=1; i<=n; i++){int u = sam.insertchar(a[i]);rt[u] = rtn; addval(rt[u],1,n,i);}sam.getord();for(int i=sam.siz; i>=1; i--){int u = sam.ord[i];rt[sam.fa[u]] = merge(rt[sam.fa[u]],rt[u]);}scanf("%d",&q);while(q--){for(int i=1; i<=samt.siz; i++) mxl[i] = 0;samt.init();scanf("%s",b+1); m = strlen(b+1); int lb,rb; scanf("%d%d",&lb,&rb);for(int i=1; i<=m; i++) b[i]-=96;for(int i=1; i<=m; i++) samt.insertchar(b[i]);samt.getord();int u = sam.rtn,uu = samt.rtn,cur = 0; llong ans = 0ll;for(int i=1; i<=m; i++){while(u && sam.son[u][b[i]]==0) {u = sam.fa[u]; cur = sam.len[u];}if(u!=0) {u = sam.son[u][b[i]]; cur++;} else {u = sam.rtn; cur = 0;}int tmp = query(rt[u],1,n,rb);while(u && tmp<lb+sam.len[sam.fa[u]]) {u = sam.fa[u]; tmp = query(rt[u],1,n,rb); cur = min(sam.len[u],tmp-lb+1);}cur = min(cur,tmp-lb+1); //注意此处必须受到区间限制 while(uu && samt.son[uu][b[i]]==0) {uu = samt.fa[uu];}uu = uu==0 ? 1 : samt.son[uu][b[i]];mxl[uu] = max(mxl[uu],cur);}for(int i=samt.siz; i>=1; i--){int uu = samt.ord[i];mxl[samt.fa[uu]] = max(mxl[samt.fa[uu]],mxl[uu]);}for(int i=1; i<=samt.siz; i++){mxl[i] = max(min(mxl[i],samt.len[i]),samt.len[samt.fa[i]]); //最长匹配的长度，不得小于该点最短长度 int tmp = samt.len[i]-mxl[i]; //该节点上不匹配的个数ans += (llong)tmp;}printf("%lld\n",ans);}return 0;
}
