SPOJ

题意：让你求这个字符串能组合出来多少个不重复的子串

思路：利用后缀数组height【i】的性质就可以求出每个位置开始能组成的字符串个数为n-sa[i]-height[i] ， n为字符串的长度

#include<iostream>#include<cstring>#include<algorithm>#include<cstdio>using namespace std;const int mx = 50005;int wa[mx],wb[mx],wv[mx],c[mx];int sa[mx],rk[mx],r[mx],height[mx];char s[mx];bool cmp(int *r,int a,int b,int i){    return r[a] == r[b]&&r[a+i] == r[b+i];}void da(int *r,int *sa,int n,int m){    int *x = wa,*y = wb;    for(int i = 0; i < m; i++)  c[i] = 0;    for(int i = 0; i < n; i++)  c[x[i]=r[i]]++;    for(int i = 1; i < m; i++)  c[i] += c[i-1];    for(int i = n-1; i >= 0; i--)   sa[--c[x[i]]] = i;    for(int j = 1,p = 1; p < n;j<<=1,m = p){        p = 0;        for(int i = n-j; i < n; i++)    y[p++] = i;        for(int i = 0; i < n; i++)  if(sa[i]>=j) y[p++] = sa[i]-j;        for(int i = 0; i < n; i++)  wv[i] = x[y[i]];        for(int i = 0; i < m; i++)  c[i] = 0;        for(int i = 0; i < n; i++)  c[wv[i]]++;        for(int i = 1; i < m; i++)  c[i] += c[i-1];        for(int i = n-1; i >= 0; i--)   sa[--c[wv[i]]] = y[i];        swap(x,y),p = 1,x[sa[0]] = 0;        for(int i = 1; i < n; i++)            x[sa[i]] = cmp(y,sa[i],sa[i-1],j)?p-1:p++;    }}void getheight(int *r,int n){    int k = 0,j;    for(int i = 1; i <= n; i++) rk[sa[i]] = i;    for(int i = 0; i < n; height[rk[i++]] = k)        for(j = sa[rk[i]-1],k?k--:0;r[i+k] == r[j+k]; k++);}int main(){    int t;    scanf("%d",&t);    while(t--){        scanf("%s",s);        int n = strlen(s);        for(int i = 0; i < n; i++)            r[i] = s[i];        r[n] = 0;        da(r,sa,n+1,256);        getheight(r,n);        int ans = 0;        for(int i = 1; i <= n; i++){            ans += n-sa[i]-height[i];            //printf("%d %d %d\n",n,sa[i],height[i]);        }        cout<<ans<<endl;    }    return 0;}