spoj705( 求不相同的子串个数)

来源：互联网发布：安卓埋雷软件编辑：程序博客网时间：2024/06/05 21:58

题意：求串s的不同子串的个数

解题思路：任何子串都是某个后缀的前缀，对n个后缀排序，求某个后缀的前缀的个数，减去height[i](第i个后缀与第i-1 个后缀有相同的height[i]个前缀)。

代码如下：

#include<iostream>#include<algorithm>#include<stdio.h>#include<math.h>#include<cstring>#include<string>#include<vector>#define N 50005#define inf 0x3f3f3f3f#define pi acos(-1.0)#define eps 10e-6using namespace std;char s[N];int r[N],wa[N],wb[N],wv[N],WS[N],sa[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 n,int m){    int i, j, p, *x = wa, *y = wb, *t;    for(i = 0; i < m; i++)  WS[i] = 0;    for(i = 0; i < n; i++)  WS[ x[i] = r[i] ]++;    for(i = 1; i < m; i++)  WS[i] += WS[i-1];    for(i = n-1; i >= 0; i--)   sa[ --WS[ 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++)  WS[i] = 0;        for(i = 0; i < n; i++)  WS[ wv[i] ]++;        for(i = 1; i < m; i++)  WS[i] += WS[i-1];        for(i = n-1; i >= 0; i--)   sa[ --WS[ 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++;    }}int rank[N],height[N];void calheight(int n){    int i,j,k = 0;    for(i = 1; i <= n; i++) rank[ sa[i] ] = i;    for(i = 0; i < n; height[ rank[i++] ] = k)        for(k ? k-- : 0, j = sa[ rank[i] - 1 ]; r[i+k] == r[j+k]; k++);}void solve(int n){    int ans = 0;    int i;  //  for(i = 1; i <= n; i++) cout<<sa[i]<<" ";cout<<endl;//    for(i = 1; i <= n; i++) cout<<height[i]<<" ";cout<<endl;    for(int i = 1; i <= n; i++)        ans += (n-sa[i]-height[i]);    printf("%d\n",ans);}int main(){    int t;    scanf("%d",&t);    while(t--)    {        int i;        scanf("%s",s);        int n = strlen(s);        for(i = 0; i < n; i++)            r[i] = s[i]+1;        r[n] = 0;        da(n+1,130);        calheight(n);    //    for(i = 0; i <= n; i++)      //      cout<<height[i]<<" ";       // cout<<endl;        solve(n);    }    return 0;}

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