Lightoj-1314 Names for Babies(后缀数组)

来源：互联网发布：自学编程入门教程编辑：程序博客网时间：2024/05/22 12:56

题目：

http://lightoj.com/volume_showproblem.php?problem=1314

题意：

求母串中长度在p到q之间的不相同子串的个数。

思路：

求不相同子串个数和明显用后缀数组可以搞，对于每个后缀能够取得子串个数就是min(q,n-sa[i])-max(p,height[i]+1)+1 很好理解就是能够取得的最大长度减去重复或者是最小长度，因为要去掉相同部分所以是取height[i]+1。

代码：

//kopyh#include <bits/stdc++.h>#include <stdio.h>#define INF 0x3f3f3f3f#define MOD 1000000007#define N 500100using namespace std;int n,m,sum,res,flag;char s[N];int seq[N], sa[N], ranks[N], height[N];int wwa[N], wwb[N], wws[N], wwv[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 n, int m){    int i, j, p, *x = wwa, *y = wwb;    for (i = 0; i < m; ++i) wws[i] = 0;    for (i = 0; i < n; ++i) wws[x[i]=r[i]]++;    for (i = 1; i < m; ++i) wws[i] += wws[i-1];    for (i = n-1; i >= 0; --i) sa[--wws[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) wwv[i] = x[y[i]];        for (i = 0; i < m; ++i) wws[i] = 0;        for (i = 0; i < n; ++i) wws[wwv[i]]++;        for (i = 1; i < m; ++i) wws[i] += wws[i-1];        for (i = n-1; i >= 0; --i) sa[--wws[wwv[i]]] = y[i];        for (swap(x, y), 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){    int i, j, k = 0;    for (i = 1; i <= n; ++i) ranks[sa[i]] = i;    for (i = 0; i < n; height[ranks[i++]] = k)        for (k?k--:0, j = sa[ranks[i]-1]; r[i+k] == r[j+k]; k++);}int main(){    int i,j,k,cas,T,t,x,y,p,q,z;    scanf("%d",&T);    cas=0;    while(T--)    {        scanf("%s",s);        scanf("%d%d",&p,&q);        n=strlen(s);m=27;        for(i=0;i<n;i++)            seq[i]=s[i]-'a'+1;        seq[n]=0;        da(seq,n+1,m);        calheight(seq,n);        res=0;        for(i=1;i<=n;i++)        {            x=max(p,height[i]);            y=min(q,n-sa[i]);            if(y-x+1>0)res+=y-x+1;        }        printf("Case %d: %d\n",++cas,res);    }    return 0;}

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