【后缀数组】 HDOJ 4552 怪盗基德的挑战书

来源：互联网发布：网络大电影盈利案例编辑：程序博客网时间：2024/06/07 16:54

题目中让我们求前缀出现的所有次数和。。。可以转化成rank【0】的后缀和其他所有后缀长度之和。。。

#include <iostream>  #include <queue>  #include <stack>  #include <map>  #include <set>  #include <bitset>  #include <cstdio>  #include <algorithm>  #include <cstring>  #include <climits>  #include <cstdlib>#include <cmath>#include <time.h>#define maxn 100005#define maxm 400005#define eps 1e-10#define mod 256#define INF 1e17#define lowbit(x) (x&(-x))  #define ls o<<1#define rs o<<1 | 1#define lson o->ch[0], L, mid  #define rson o->ch[1], mid+1, R  typedef long long LL;//typedef int LL;using namespace std;char s[maxn];int sa[maxn], t[maxn], t2[maxn], c[maxn];int height[maxn], rank[maxn];/*void init(void){cc[0] = 0;for(int i = 1; i < maxn; i++)cc[i] = (cc[i-1] + i) % mod;}*/void build(int n, int m){int i, *x = t, *y = t2, k, p;for(i = 0; i < m; i++) c[i] = 0;for(i = 0; i < n; i++) c[x[i] = s[i]]++;for(i = 1; i < m; i++) c[i] += c[i-1];for(i = n-1; i >= 0; i--) sa[--c[x[i]]] = i;for(k = 1; k <= n; k <<= 1) {p = 0;for(i = n-k; i < n; i++) y[p++] = i;for(i = 0; i < n; i++) if(sa[i] >= k) y[p++] = sa[i] - k;for(i = 0; i < m; i++) c[i] = 0;for(i = 0; i < n; i++) c[x[y[i]]]++;for(i = 1; i < m; i++) c[i] += c[i-1];for(i = n-1; i >= 0; i--) sa[--c[x[y[i]]]] = y[i];swap(x, y), p = 1, x[sa[0]] = 0;for(i = 1; i < n; i++)x[sa[i]] = y[sa[i]] == y[sa[i-1]] && y[sa[i]+k] == y[sa[i-1]+k] ? p-1 : p++;if(p >= n) break;m = p;}}void getheight(int n){int i, j, k = 0;for(i = 0; i <= n; i++) rank[sa[i]] = i;for(i = 0; i < n; i++) {if(k) k--;j = sa[rank[i]-1];while(s[i+k] == s[j+k]) k++;height[rank[i]] = k;}}void work(int n){int ans = 0, tmp, now;now = rank[0];tmp = height[now+1];for(int i = now+1; i <= n; i++) {tmp = min(tmp, height[i]);ans = (ans + tmp) % mod;}tmp = height[now];for(int i = now; i >= 1; i--) {tmp = min(tmp, height[i]);ans = (ans + tmp) % mod;}ans += n;ans %= mod;printf("%d\n", ans);}void debug(int n){for(int i = 1; i <= n; i++)printf("%d\n", height[i]);}int main(void){int n;while(scanf("%s", s)!=EOF) {n = strlen(s);build(n+1, 128);getheight(n);work(n);}return 0;}

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