【后缀数组】 HDOJ 3518 Boring counting

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后缀数组的应用。。先求出后缀数组的sa数组和height数组，显然，求height中发现在名次数组中就可以看出通过名次能把后缀数组分成好几份，每一份都可能包含可行的子串。。然后枚举子串长度。。然后每次在已经按名次分好的height数组里找，对于已经分好的后缀数组的每一份，如果这一份的所有值通过sa映射的最大值减去最小值大于等于当前枚举的长度，那么这一份后缀数组中必然包含当前长度的子串，那么ans+1。显然每一份的后缀数组和其他份的必然不重合。最后把答案输出就好了。。。

#include <iostream>  #include <sstream>  #include <algorithm>  #include <vector>  #include <queue>  #include <stack>  #include <map>  #include <set>  #include <bitset>  #include <cstdio>  #include <cstring>  #include <cstdlib>  #include <cmath>  #include <climits>  #define maxn 100005#define eps 1e-6 #define mod 10007 #define INF 99999999  #define lowbit(x) (x&(-x))  //#define lson o<<1, L, mid  //#define rson o<<1 | 1, mid+1, R  typedef long long LL;using namespace std;char s[maxn];int t[maxn], t2[maxn], c[maxn], sa[maxn];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;}}int rank[maxn], height[maxn];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 debug(int n){int i;for(i = 0; i < n; i++)printf("%d   %d\n", height[i], i);}int solve(int p, int n){int ans = 0, _min, _max, i;_min = INF, _max = 0;for(i = 1; i <= n; i++) {if(height[i] >= p) {_max = max(_max, sa[i]);_min = min(_min, sa[i]);_max = max(_max, sa[i-1]);_min = min(_min, sa[i-1]);}else {if(_max - _min >= p) ans++;_max = 0;_min = INF;}}if(_max - _min >= p) ans++;return ans;}void work(void){int n = strlen(s), ans, p;build(n+1, 128);getheight(n);ans = 0;for(p = 1; p <= (n>>1); p++)ans += solve(p, n);printf("%d\n", ans);}int main(void){while(scanf("%s", s)!=EOF) {if(s[0] == '#') break;work();}return 0;}

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