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SPOJ 694 705 后缀数组

后缀数组求不同子序列数量的简单题。

对于一个已经处理好的后缀数组,对于每一个suffix(i),必然会产生len - sa[i]个前缀(假设从0开始), 然后这len - sa[i]个前缀里面有height[i]个是和前面那个前缀相同的,所以是len - sa[i] - height[i]个,求一下和就好。。

#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <stack>#include <map>#include <set>#include <climits>#include <string>using namespace std; #define MP make_pair#define PB push_backtypedef long long LL;typedef unsigned long long ULL;typedef vector<int> VI;typedef pair<int, int> PII;typedef pair<double, double> PDD;const int INF = INT_MAX / 3;const double eps = 1e-8;const LL LINF = 1e17;const double DINF = 1e60;const int maxn = 50005;//以下是倍增法求后缀数组int wa[maxn], wb[maxn], wv[maxn], ws[maxn];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 *sa, 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 <<= 1, 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 = 0; 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++;    }}//以下是求解height 数组int height[maxn], Rank[maxn];void calheight(int *r, int *sa, 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++) ;}char buf[maxn];int len, str[maxn], sa[maxn];int main() {    int T; scanf("%d", &T);    while(T--) {        scanf("%s", buf);        len = strlen(buf);        for(int i = 0; i < len; i++) str[i] = buf[i] + 1;        str[len] = 0;        da(str, sa, len + 1, 200);        calheight(str, sa, len);        int ans = 0;        for(int i = 1; i <= len; i++) {            ans += len - sa[i] - height[i];        }        printf("%d\n", ans);    }    return 0;}

 

SPOJ 694 705 后缀数组