Luogu3732 [HAOI2017] 供给侧改革 【后缀数组】【线段树】【乱搞】
题目分析：
这道题我是乱搞的，因为他说$01$串是随机的。
那么我们可以猜测能够让LCP变大的地方很少。求出后缀数组之后可能让LCP变大的地方就等价于从大到小往height里动态加点同时维护这个点左右两边的单调栈。
这个事情用线段树模拟就行了。
用暴力跑一下发现果然不大，只有200w，其中很多还是废点，把废点特判一下删除掉就行了。
然后把询问离线，按r排序，再作扫描线，每次扫到r的时候把前面变动的LCP更新一下。然后对于询问也做一个类似上面单调栈的操作。
代码：
1 #include<bits/stdc++.h> 2 #define RI register 3 using namespace std; 4 5 const int maxn = 100020; 6 const int N = 100000; 7 8 int n,q; 9 char str[maxn]; 10 int T[maxn<<2],a[maxn]; 11 int RMQ[maxn][19],pw[maxn]; 12 13 struct node{int l,r,nb;}qy[maxn]; 14 int ans[maxn]; 15 int sa[maxn],rk[maxn],X[maxn],Y[maxn]; 16 int ht[maxn],h[maxn]; 17 18 void fstread(int &x){ 19 char ch = getchar();x = 0; 20 while(ch > '9' || ch < '0') ch = getchar(); 21 while(ch >= '0' && ch <= '9') x = x*10+ch-'0',ch=getchar(); 22 } 23 24 int chk(int x,int k){ 25 return rk[sa[x]]==rk[sa[x-1]]&&rk[sa[x]+(1<<k)]==rk[sa[x-1]+(1<<k)]; 26 } 27 28 void GetSA(){ 29 for(RI int i=0;i<n;i++) X[str[i]]++; 30 for(RI int i=1;i<=N;i++) X[i] += X[i-1]; 31 for(RI int i=n-1;i>=0;i--) sa[X[str[i]]--] = i; 32 for(RI int i = 2, num = 1;i <= n;i++) 33 rk[sa[i]] = (str[sa[i]] == str[sa[i-1]]?num:++num); 34 rk[sa[1]] = 1; 35 for(int k=1;(1<<k-1)<=n;k++){ 36 for(RI int i=1;i<=N;i++) X[i] = 0; 37 for(RI int i=n-(1<<k-1);i<n;i++) Y[i-n+(1<<k-1)+1]=i; 38 for(RI int i=1,j=(1<<k-1)+1;i<=n;i++) 39 if(sa[i]>=(1<<k-1))Y[j++]=sa[i]-(1<<k-1); 40 for(RI int i=0;i<n;i++) X[rk[i]]++; 41 for(RI int i=1;i<=N;i++) X[i]+=X[i-1]; 42 for(RI int i=n;i>=1;i--) sa[X[rk[Y[i]]]--] = Y[i]; 43 int num = 1; Y[sa[1]] = 1; 44 for(RI int i=2;i<=n;i++) Y[sa[i]] = (chk(i,k-1)?num:++num); 45 for(RI int i=0;i<n;i++) rk[i] = Y[i]; 46 if(num == n) break; 47 } 48 } 49 50 void buildRMQ(){ 51 for(int i=1;i<=n;i++) RMQ[i][0] = ht[i]; 52 for(int i=0;(1<<i)<=n;i++) pw[(1<<i)] = i; 53 for(int i=3;i<=n;i++) if(!pw[i]) pw[i] = pw[i-1]; 54 for(int k=1;(1<<k)<=n;k++){ 55 for(RI int i=1;i<=n;i++){ 56 if(i+(1<<k-1)>n) RMQ[i][k] = RMQ[i][k-1]; 57 else RMQ[i][k] = min(RMQ[i][k-1],RMQ[i+(1<<k-1)][k-1]); 58 } 59 } 60 } 61 62 void GetHeight(){ 63 for(RI int i=0;i<n;i++){ 64 if(i) h[i] = max(0,h[i-1]-1); 65 if(rk[i] == 1) {h[i] = 0;continue;} 66 while(str[i+h[i]] == str[sa[rk[i]-1]+h[i]]) h[i]++; 67 } 68 for(RI int i=0;i<n;i++) ht[rk[i]] = h[i]; 69 } 70 71 int LCP(int l,int r){ 72 l = rk[l],r = rk[r]; if(l > r) swap(l,r);l++; 73 int k = pw[r-l+1]; 74 return min(RMQ[l][k],RMQ[r-(1<<k)+1][k]); 75 } 76 77 void update(int now,int tl,int tr,int pos,int len){ 78 if(tl == tr){T[now] = max(T[now],len); return;} 79 int mid = (tl+tr)/2; 80 if(pos <= mid) update(now<<1,tl,mid,pos,len); 81 else update(now<<1|1,mid+1,tr,pos,len); 82 T[now] = max(T[now<<1],T[now<<1|1]); 83 } 84 85 int Query(int now,int tl,int tr,int l,int r,int mx){ 86 if(tl >= l && tr <= r){ 87 if(T[now] <= mx) return -1; 88 else{ 89 while(tl < tr){ 90 int mid = (tl+tr)/2; 91 if(T[now<<1|1] > mx) tl = mid+1,now = now*2+1; 92 else tr = mid,now<<=1; 93 } 94 return tl; 95 } 96 } 97 if(tl > r || tr < l) return -1; 98 int mid = (tl+tr)/2; 99 if(mid >= r) return Query(now<<1,tl,mid,l,r,mx); 100 if(mid < l) return Query(now<<1|1,mid+1,tr,l,r,mx); 101 int z = Query(now<<1|1,mid+1,tr,l,r,mx); 102 if(z!=-1) return z; 103 else return Query(now<<1,tl,mid,l,r,mx); 104 } 105 106 vector<int> mlgb[maxn]; 107 108 namespace BruteForce{ 109 int minn[maxn<<2]; 110 111 pair<int,int> QueryRange(int now,int tl,int tr,int l,int r,int up,int d){ 112 if(tl >= l && tr <= r && d == 1){ 113 if(minn[now] < up){ 114 while(tl != tr){ 115 int mid = (tl+tr)/2; 116 if(minn[now<<1|1] < up) tl = mid+1,now=now*2+1; 117 else tr = mid,now<<=1; 118 } 119 return make_pair(minn[now],tl); 120 }else return make_pair(0,0); 121 } 122 if(tl >= l && tr <= r && d == 0){ 123 if(minn[now] < up){ 124 while(tl != tr){ 125 int mid = (tl+tr)/2; 126 if(minn[now<<1] < up) tr = mid,now<<=1; 127 else tl = mid+1,now=now*2+1; 128 } 129 return make_pair(minn[now],tl); 130 }else return make_pair(0,0); 131 } 132 if(tl > r || tr < l) return make_pair(0,0); 133 int mid = (tl+tr)/2; 134 if(mid>=r) return QueryRange(now<<1,tl,mid,l,r,up,d); 135 if(mid<l) return QueryRange(now<<1|1,mid+1,tr,l,r,up,d); 136 if(d){ 137 pair<int,int> z = QueryRange(now<<1|1,mid+1,tr,l,r,up,d); 138 if(z.first) return z; 139 else return QueryRange(now<<1,tl,mid,l,r,up,d); 140 }else{ 141 pair<int,int> z = QueryRange(now<<1,tl,mid,l,r,up,d); 142 if(z.first) return z; 143 else return QueryRange(now<<1|1,mid+1,tr,l,r,up,d); 144 } 145 } 146 void add(int now,int tl,int tr,int pos,int dt){ 147 if(tl == tr) {minn[now] = dt;return;} 148 int mid = (tl+tr)/2; 149 if(pos <= mid) add(now<<1,tl,mid,pos,dt); 150 else add(now<<1|1,mid+1,tr,pos,dt); 151 minn[now] = min(minn[now<<1],minn[now<<1|1]); 152 } 153 154 void holyshit(){ 155 memset(minn,0x3f,sizeof(minn)); 156 for(int i=n-1;i>=0;i--){ 157 int z = rk[i],p=19260817; 158 add(1,1,n,z,i); 159 while(true){ 160 pair<int,int> um = QueryRange(1,1,n,z+1,n,p,0); 161 if(!um.first) break; 162 mlgb[um.first].push_back(i); 163 z = um.second; p = sa[z]; 164 } 165 z = rk[i],p = 19260817; 166 while(true){ 167 pair<int,int> um = QueryRange(1,1,n,1,z-1,p,1); 168 if(!um.first) break; 169 mlgb[um.first].push_back(i); 170 z = um.second; p = sa[z]; 171 } 172 } 173 174 } 175 } 176 177 void work(){ 178 GetSA(); 179 GetHeight(); 180 buildRMQ(); 181 182 BruteForce::holyshit(); 183 int num = 1; 184 for(int i=0;i<n;i++){ 185 for(int j=0;j<mlgb[i].size();j++){ 186 int z = LCP(mlgb[i][j],i); 187 if(a[mlgb[i][j]] >= z) continue; 188 a[mlgb[i][j]] = z; 189 update(1,0,n-1,mlgb[i][j],z); 190 191 } 192 while(qy[num].r == i){ 193 int z = 0,pq = qy[num].r-1; 194 while(pq >= qy[num].l){ 195 int hp = Query(1,0,n-1,0,pq,z); 196 if(hp < qy[num].l) hp = qy[num].l-1; 197 ans[qy[num].nb] += (pq-hp)*z; 198 if(hp < 0) break; 199 z = a[hp],pq = hp; 200 } 201 num++; 202 } 203 } 204 for(int i=1;i<=q;i++) printf("%d\n",ans[i]); 205 } 206 207 int cmp(node alpha,node beta){return alpha.r < beta.r;} 208 int main(){ 209 //freopen("1.in","r",stdin); 210 fstread(n);fstread(q); 211 for(int i=1;i<=n;i++){ 212 str[i-1] = getchar(); 213 while(str[i-1]!= '0' && str[i-1] != '1') str[i-1] = getchar(); 214 } 215 for(int i=1;i<=q;i++) fstread(qy[i].l),fstread(qy[i].r),qy[i].nb=i; 216 for(int i=1;i<=q;i++) qy[i].l--,qy[i].r--; 217 sort(qy+1,qy+q+1,cmp); 218 work(); 219 return 0; 220 }
