标签:read tchar dfs main 计算 build c++ can 后缀
先上代码
#include <bits/stdc++.h> /* a.长度不一定刚好是len*2-2,所以要计算kmp的长度 b.忘了返回值 */ using namespace std; const int N = 101000,LOGN = 20,M = 301000; char read(){ char ch = getchar(); while (ch < ‘a‘ || ‘z‘ < ch) ch = getchar(); return ch; } struct Edge{ int next,end; }edge[LOGN*N]; struct Edge2{ int next,end;char ch; }edge2[N<<1]; struct Node{ int cnt,son[2]; }nod[N*LOGN]; char ch[N],s[M],d[M]; int efn,fa[N][LOGN],first[N][LOGN];//倍增 int first2[N],efn2; int sa[N],rank[N];//后缀数组 sa 排名为i的是 rank->i的排名 int dep[N],siz[N],hson[N],fat[N],top[N];//树剖 int next[N];//kmp int root[N],Len; int n,m,o,lg2[N],len; void init(); void addedge(int,int,int); void addedge(int,int,char); void dfs(int,int); void dfs3(int,int); void dfs2(int,int); void insert(int,int,int,int,int); void build(int,int,int); int lca(int,int); int ef1(int,int); int ef2(int,int); int getans(int,int,int,int,int,int); bool check1(int); bool check2(int); int main(){ scanf("%d%d",&n,&m); lg2[0] = -1; for (int i = 1;i <= n;i++) lg2[i] = lg2[i/2]+1; o = lg2[n]+1; for (int i = 1;i < n;i++){ int x,y;char ch; scanf("%d%d",&x,&y); ch = read(); addedge(x,y,ch); } dfs(1,0); top[1] = 1;dfs3(1,0); for (int i = 1;i <= n;i++) if (fa[i][0]) addedge(fa[i][0],i,0); for (int i = 1;i <= o;i++){ for (int j = 1;j <= n;j++){ fa[j][i] = fa[fa[j][i-1]][i-1]; if (fa[j][i]) addedge(fa[j][i],j,i); } } init(); len = 1;root[0] = 1;build(1,1,n); dfs2(1,0); ch[1] = ‘ ‘; for (int i = 1;i <= m;i++){ int x,y,z,u,v; int l,r,ans = 0,len,cntt = 0; scanf("%d%d",&x,&y);z = lca(x,y); scanf("%s",s);len = strlen(s);Len = len; u = x;v = y; for (int j = 16;j >= 0;j--){ if (dep[fa[u][j]] >= dep[z]+len-1) u = fa[u][j]; if (dep[fa[v][j]] >= dep[z]+len-1) v = fa[v][j]; } cntt = dep[u]-dep[z] + dep[v] - dep[z];//a l = ef1(2,n+1); r = ef2(1,n); if (l <= r)ans += getans(root[u],root[x],1,n,min(n,l),max(2,r)); reverse(s+0,s+len); l = ef1(2,n+1); r = ef2(1,n); if (l <= r)ans += getans(root[v],root[y],1,n,min(n,l),max(2,r)); int t1 = 0,w1 = cntt-1; while (u != z){ d[t1++] = ch[u]; u = fat[u]; } while (v != z){ d[w1--] = ch[v]; v = fat[v]; } reverse(s+0,s+len); w1 = cntt; next[0] = -1;t1 = -1; for (int j = 1;j < len;j++){ t1++; while (t1 && s[t1] != s[j]) t1 = next[t1]; if (s[t1] == s[j]) next[j] = t1; else next[j] = --t1; } int now = -1; for (int j = 0;j < w1;j++){ while (now != -1 && s[now+1] != d[j]) now = next[now]; if (s[now+1] == d[j]) now++; if (now == len-1) {ans++;now = next[now];} } printf("%d\n",ans); } return 0; } void init(){ static int x2[N],y2[N],a[N]; int *x = x2,*y = y2,m = 256,cnt = -1; for (int i = 0;i <= m;i++) a[i] = 0; for (int i = 1;i <= n;i++) a[x[i] = ch[i]]++; for (int i = 1;i <= m;i++) a[i] += a[i-1]; for (int i = 1;i <= n;i++) sa[a[ch[i]]--] = i; x[0] = -1; for (int k = 1;k <= n;k <<= 1){ int p = 0;cnt++; for (int i = 0;i <= m;i++) a[i] = 0; for (int i = 1;i <= n;i++) if (fa[sa[i]][cnt] <= 1) y[++p] = sa[i]; for (int i = 2;i <= n;i++) for (int h = first[sa[i]][cnt];h;h = edge[h].next){ int u = edge[h].end; y[++p] = u; } for (int i = 1;i <= n;i++) a[x[y[i]]]++; for (int i = 1;i <= m;i++) a[i] += a[i-1]; for (int i = n;i >= 1;i--) sa[a[x[y[i]]]--] = y[i]; swap(x,y); p = 1; x[sa[1]] = p; for (int i = 2;i <= n;i++){ int u = fa[sa[i]][cnt],v = fa[sa[i-1]][cnt]; if (u == 1) u = 0;if (v == 1) v = 0; x[sa[i]] = (y[sa[i]] == y[sa[i-1]]) ? (u == 0 && v == 0 ? p : y[u] == y[v] ? p : ++p) : ++p; } m = p; if (m >= n) break; } for (int i = 1;i <= n;i++) rank[sa[i]] = i; } void addedge(int x,int y,int z){ edge[++efn].end = y; edge[ efn].next = first[x][z]; first[x][z] = efn; } void dfs(int x,int y){ fa[x][0] = y; siz[x] = 1;fat[x] = y;dep[x] = dep[y]+1; for (int h = first2[x];h;h = edge2[h].next){ int u = edge2[h].end; if (u != y) { ch[u] = edge2[h].ch; dfs(u,x); siz[x] += siz[u]; hson[x] = siz[u] > siz[hson[x]] ? u : hson[x]; } } } void dfs3(int x,int y){ if (hson[x]){ top[hson[x]] = top[x]; dfs3(hson[x],x); } for (int h = first2[x];h;h = edge2[h].next){ int u = edge2[h].end; if (u != y && u != hson[x]){ top[u] = u; dfs3(u,x); } } } void dfs2(int x,int y){ root[x] = ++len; insert(root[y],root[x],1,n,rank[x]); for (int h = first[x][0];h;h = edge[h].next){ int u = edge[h].end; dfs2(u,x); } } void build(int p,int l,int r){ if (l == r) return; int mid = l + r >> 1; nod[p].son[0] = ++len; nod[p].son[1] = ++len; build(nod[p].son[0],l,mid); build(nod[p].son[1],mid+1,r); } void insert(int p,int q,int l,int r,int x){ nod[q].cnt = nod[p].cnt+1; if (l == r) return; int mid = l + r >> 1; if (x <= mid) { nod[q].son[0] = ++len; nod[q].son[1] = nod[p].son[1]; insert(nod[p].son[0],nod[q].son[0],l,mid,x); } else{ nod[q].son[1] = ++len; nod[q].son[0] = nod[p].son[0]; insert(nod[p].son[1],nod[q].son[1],mid+1,r,x); } } int lca(int x,int y){ while (top[x] != top[y]){ if (dep[top[x]] < dep[top[y]]) y = fat[top[y]]; else x = fat[top[x]]; } return dep[x] < dep[y] ? x : y; } void addedge(int x,int y,char ch){ edge2[++efn2].end = y; edge2[ efn2].ch = ch; edge2[ efn2].next = first2[x]; first2[x] = efn2; edge2[++efn2].end = x; edge2[ efn2].ch = ch; edge2[ efn2].next = first2[y]; first2[y] = efn2; } int ef1(int l,int r){ int mid = l + r >> 1; while (l < r){ if (check1(sa[mid])) r = mid;else l = mid+1; mid = l + r >> 1; } return l; } int ef2(int l,int r){ int mid = l + r + 1 >> 1; while (l < r){ if (check2(sa[mid])) l = mid;else r = mid-1; mid = l + r + 1 >> 1; } return l; } bool check1(int p){ for (int i = 0;i < Len;i++){ if (p == 1) return 0; if (ch[p] < s[i]) return 0; if (ch[p] > s[i]) return 1; p = fat[p]; } return 1; } bool check2(int p){ for (int i = 0;i < Len;i++){ if (p == 1) return 1; if (ch[p] < s[i]) return 1; if (ch[p] > s[i]) return 0; p = fat[p]; } if (p == 1) return 1; return 1;//b } int getans(int p,int q,int l,int r,int x,int y){ if (y < x) return 0; if (l == x && r == y) return nod[q].cnt - nod[p].cnt; int mid = l + r >> 1; if (y <= mid) return getans(nod[p].son[0],nod[q].son[0],l,mid,x,y); else if (mid < x) return getans(nod[p].son[1],nod[q].son[1],mid+1,r,x,y); else return getans(nod[p].son[0],nod[q].son[0],l,mid,x,mid)+getans(nod[p].son[1],nod[q].son[1],mid+1,r,mid+1,y); }
标签:read tchar dfs main 计算 build c++ can 后缀
原文地址:http://www.cnblogs.com/victbr/p/6510725.html