BZOJ3230 相似子串 【后缀数组】
题目分析：
容易想到sa排好序之后，子串排名就是前面的子串减去height数组。所以正着做一遍，倒着做一遍就行了。
代码：
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 const int maxn = 252000; 5 const int N = 100000; 6 7 int n,q; 8 char str[maxn]; 9 10 int sa[maxn],rk[maxn],X[maxn],Y[maxn]; 11 int height[maxn],h[maxn],RMQ[maxn][19]; 12 int len[maxn],st1[maxn],st2[maxn]; 13 long long stnum[maxn]; 14 15 int chk(int x,int k){ 16 return rk[sa[x]]==rk[sa[x-1]]&&rk[sa[x]+(1<<k)]==rk[sa[x-1]+(1<<k)]; 17 } 18 19 void getsa(){ 20 for(int i=0;i<n;i++) X[str[i]]++; 21 for(int i=1;i<=N;i++) X[i] += X[i-1]; 22 for(int i=n-1;i>=0;i--) sa[X[str[i]]--] = i; 23 for(int i = 2, num = 1;i <= n;i++) 24 rk[sa[i]] = (str[sa[i]] == str[sa[i-1]]?num:++num); 25 rk[sa[1]] = 1; 26 for(int k=1;(1<<k-1)<=n;k++){ 27 for(int i=1;i<=N;i++) X[i] = 0; 28 for(int i=n-(1<<k-1);i<n;i++) Y[i-n+(1<<k-1)+1]=i; 29 for(int i=1,j=(1<<k-1)+1;i<=n;i++) 30 if(sa[i]>=(1<<k-1))Y[j++]=sa[i]-(1<<k-1); 31 for(int i=0;i<n;i++) X[rk[i]]++; 32 for(int i=1;i<=N;i++) X[i]+=X[i-1]; 33 for(int i=n;i>=1;i--) sa[X[rk[Y[i]]]--] = Y[i]; 34 int num = 1; Y[sa[1]] = 1; 35 for(int i=2;i<=n;i++) Y[sa[i]] = (chk(i,k-1)?num:++num); 36 for(int i=0;i<n;i++) rk[i] = Y[i]; 37 if(num == n) break; 38 } 39 } 40 41 void getheight(){ 42 for(int i=0;i<n;i++){ 43 if(i) h[i] = max(0,h[i-1]-1); else h[i] = 0; 44 if(rk[i] == 1) continue; 45 int comp = sa[rk[i]-1]; 46 while(str[comp+h[i]] == str[i+h[i]])h[i]++; 47 } 48 for(int i=0;i<n;i++) height[rk[i]] = h[i]; 49 for(int i=1;i<=n;i++) RMQ[i][0] = height[i]; 50 for(int k=1;(1<<k)<=n;k++){ 51 for(int i=1;i<=n;i++){ 52 if(i+(1<<k-1)>n) RMQ[i][k] = RMQ[i][k-1]; 53 else RMQ[i][k] = min(RMQ[i][k-1],RMQ[i+(1<<k-1)][k-1]); 54 } 55 } 56 } 57 58 int getLCP(int L,int R){ 59 if(L > R) swap(L,R); 60 if(L == R) return n-sa[L]; 61 L++; 62 int k = 0; while((1<<k+1)<=R-L+1)k++; 63 return min(RMQ[L][k],RMQ[R-(1<<k)+1][k]); 64 } 65 66 void read(){ 67 scanf("%d%d",&n,&q); 68 scanf("%s",str); 69 } 70 71 long long ans[maxn]; 72 void work(){ 73 getsa(); 74 getheight(); 75 for(int i=1;i<=n;i++) stnum[i] = stnum[i-1]+(n-sa[i])-height[i]; 76 for(int i=1;i<=q;i++){ 77 long long u,v; scanf("%lld%lld",&u,&v); 78 if(u > v) swap(u,v); if(v > stnum[n]){ans[i] = -1;continue;} 79 int pn = lower_bound(stnum+1,stnum+n+1,u)-stnum; 80 int pm = lower_bound(stnum+1,stnum+n+1,v)-stnum; 81 len[i] = min(height[pn]+u-stnum[pn-1],height[pm]+v-stnum[pm-1]); 82 st1[i] = n-(sa[pn]+height[pn]+u-stnum[pn-1]); 83 st2[i] = n-(sa[pm]+height[pm]+v-stnum[pm-1]); 84 int LCP = getLCP(pn,pm); LCP = min(LCP,len[i]); 85 ans[i]+=1ll*LCP*LCP; 86 } 87 for(int i=0;i<n/2;i++) swap(str[i],str[n-i-1]); 88 memset(sa,0,sizeof(sa)); 89 memset(rk,0,sizeof(rk)); 90 memset(X,0,sizeof(X)); 91 memset(Y,0,sizeof(Y)); 92 memset(height,0,sizeof(height)); 93 memset(h,0,sizeof(h)); 94 memset(RMQ,0,sizeof(RMQ)); 95 getsa(); 96 getheight(); 97 for(int i=1;i<=q;i++){ 98 if(ans[i] == -1) continue; 99 int LCP = getLCP(rk[st1[i]],rk[st2[i]]); LCP = min(LCP,len[i]); 100 ans[i] += 1ll*LCP*LCP; 101 } 102 for(int i=1;i<=q;i++) printf("%lld\n",ans[i]); 103 } 104 105 int main(){ 106 read(); 107 work(); 108 return 0; 109 }
