后缀数组模板

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#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>using namespace std;const int maxn = 1000;int rank[maxn],wb[maxn],wv[maxn],wss[maxn];int n;bool cmp(int *r,int a,int b,int l){    return r[a]==r[b] && r[a+l]==r[b+l];}void da(int *r,int *sa,int n,int m){    int i,j,p,*x=rank,*y=wb,*t;    for(i=0;i<m;i++) wss[i]=0;    for(i=0;i<n;i++) wss[x[i]=r[i]]++;    for(i=1;i<m;i++) wss[i]+=wss[i-1];    for(i=n-1;i>=0;i--)       sa[--wss[x[i]]]=i;    for(j=1,p=1;p<n;j*=2,m=p)    {        for(p=0,i=n-j;i<n;i++) y[p++]=i;        for(i=0;i<n;i++) if(sa[i]>=j) y[p++]=sa[i]-j;        for(i=0;i<n;i++) wv[i]=x[y[i]];        for(i=0;i<m;i++) wss[i]=0;        for(i=0;i<n;i++) wss[wv[i]]++;        for(i=1;i<m;i++) wss[i]+=wss[i-1];        for(i=n-1;i>=0;i--) sa[--wss[wv[i]]]=y[i];        for(t=x,x=y,y=t,p=1,x[sa[0]]=0,i=1;i<n;i++)        x[sa[i]]=cmp(y,sa[i-1],sa[i],j)?p-1:p++;    }    return;}int main(){    int a;    a = 2;    char s[1000],l;    int r[1000],sa[1000],i;    while(scanf("%s",s) == 1)    {        puts(s);        l = strlen(s);   l++;        for(i=0; i<l-1; i++) r[i] = s[i]-'a'+1;        r[l-1] = 0;        da(r,sa,l,27);        //------------------------------------------        for(i=0; i<l-1; i++)  // rank[i] : suffix(i)排第几           printf("rank[%d] =  %d\n",i,rank[i]);        printf("\n");        for(i=0; i<l; i++)   // sa[i] : 排在第i个的是谁           printf("sa[%d] =  %d\n",i,sa[i]);        //------------------------------------------    }    return 0;}

求height[i]:相邻后缀的最长公共前缀的模板

POJ3693:求重复次数最多的连续重复子串。(并且含有RMQ，Log数组尤其给力）

#include<iostream>#include<cstdio>#include<vector>#include<cstring>using namespace std;const int nMax =1000012;int  num[nMax];int sa[nMax], rank[nMax], height[nMax];int wa[nMax], wb[nMax], wv[nMax], wd[nMax];int cmp(int *r, int a, int b, int l){    return r[a] == r[b] && r[a+l] == r[b+l];}void da(int *r, int n, int m){          //  倍增算法 r为待匹配数组  n为总长度 m为字符范围    int i, j, p, *x = wa, *y = wb, *t;    for(i = 0; i < m; i ++) wd[i] = 0;    for(i = 0; i < n; i ++) wd[x[i]=r[i]] ++;    for(i = 1; i < m; i ++) wd[i] += wd[i-1];    for(i = n-1; i >= 0; i --) sa[-- wd[x[i]]] = i;    for(j = 1, p = 1; p < n; j *= 2, m = p){        for(p = 0, i = n-j; i < n; i ++) y[p ++] = i;        for(i = 0; i < n; i ++) if(sa[i] >= j) y[p ++] = sa[i] - j;        for(i = 0; i < n; i ++) wv[i] = x[y[i]];        for(i = 0; i < m; i ++) wd[i] = 0;        for(i = 0; i < n; i ++) wd[wv[i]] ++;        for(i = 1; i < m; i ++) wd[i] += wd[i-1];        for(i = n-1; i >= 0; i --) sa[-- wd[wv[i]]] = y[i];        for(t = x, x = y, y = t, p = 1, x[sa[0]] = 0, i = 1; i < n; i ++){            x[sa[i]] = cmp(y, sa[i-1], sa[i], j) ? p - 1: p ++;        }    }}void calHeight(int *r, int n){           //  求height数组。    int i, j, k = 0;    for(i = 1; i <= n; i ++) rank[sa[i]] = i; // 1->n    for(i = 0; i < n; i++){        for(k ? k -- : 0, j = sa[rank[i]-1]; r[i+k] == r[j+k]; k ++);        height[rank[i]] = k;    } /*   for(i=0; i<n; i++)       printf("i=%d height[i]=%d\n",i,height[i]); */}int Log[nMax];int best[20][nMax];void initRMQ(int n) {//初始化RMQfor(int i = 1; i <= n ; i ++) best[0][i] = height[i];for(int i = 1; i <= Log[n] ; i ++) {int limit = n - (1<<i) + 1;for(int j = 1; j <= limit ; j ++) {best[i][j] = min(best[i-1][j] , best[i-1][j+(1<<i>>1)]);}}}int lcp(int a,int b) {//询问a,b后缀的最长公共前缀a = rank[a];    b = rank[b];if(a > b) swap(a,b);a ++;int t = Log[b - a + 1];return min(best[t][a] , best[t][b - (1<<t) + 1]);}char str[nMax];int ans[nMax];int main(){    int i,j,n,cas=0;Log[0] = -1;for(int i=1;i<=nMax;i++){ // 求log2,这么强大的位运算。。Log[i]=(i&(i-1))?Log[i-1]:Log[i-1] + 1 ;/*if(i <= 40)  printf("i=%d Log[i]=%d\n",i,Log[i]);*/}    while(scanf("%s",str)!=EOF&&str[0]!='#'){        n=strlen(str);        /*----------------------------*/        for(i=0;i<n;i++){            num[i]=str[i]-'a'+1;        }num[n]=0;        da(num, n + 1,30);    /*    for(i=0; i<=n; i++) // <= not <!!!           printf("i=%d sa[i]=%d\n",i,sa[i]); */        /*----------------------------*/        calHeight(num,n);        initRMQ(n);        int l,r,t,k,maxx=-1,a;        for(l=1;l<n;l++){     //枚举长度            for(i=0;i+l<n;i+=l){                k=lcp(i,i+l);                r=k/l+1;                t=l-k%l;                t=i-t;                if (t>=0&&k%l!=0){                    if(lcp(t,t+l)>=k) r++;                }                if(r>maxx){                    a=0;                    maxx=r;                    ans[a++]=l;                }                if(a==maxx){                    ans[a++]=l;                }            }        }        int start,b=0,c;        for(i=1;i<=n&&!b;i++){  //sa数组是保证字典序的神器啊 Orz            for(j=0;j<a;j++){                int tl=ans[j];                if(lcp(sa[i],sa[i]+tl)>=(maxx-1)*tl){                    start=sa[i];                    l=tl*maxx;                    b=1;                    break;                }            }        }        printf("Case %d: ",++cas);        for (int i=0;i<l;i++) printf("%c",str[start+ i]);        printf("\n");    }    return 0;}