Luogu3732 [HAOI2017] 供给侧改革 【后缀数组】【线段树】【乱搞】
作者：互联网
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 }
标签：RMQ,num,int,tr,乱搞,um,HAOI2017,Luogu3732,now 来源： https://www.cnblogs.com/Menhera/p/10555157.html