题意就是求一个字符串的重复出现（出现次数>=2）的不同子串的个数。
标准解法是后缀数组、最长公共前缀的应用，对于样例aabaab，先将所有后缀排序：
aab
3 aabaab
1 ab
2 abaab
0 b
1 baab
每个后缀前面数字代表这个后缀与它之前的后缀（rank比它小1）的最长公共前缀的长度：然而就可以这样理解这个最长公共前缀LCP、aabaab与aab的最长公共前缀是3，那说明子串a、aa、aab都至少出现的两次，那么这就是后缀aab重复出现的子串个数
然后我们考虑后缀ab与后缀aabaab的最长公共前缀=1，这时由于LCP(ab, aabaab) <= LCP(aabaab, aab)，所以就说明ab与aabaab的所有公共前缀(只有LCP(ab, aabaab) = 1个)之前都已经计算过了，所以我们直接跳过
之后我们考虑后缀abaab与后缀ab的最长公共前缀LCP(abaab, ab) > LCP(ab, aabaab),同时，由于LCP(ab, aabaab) != 0，所以考虑abaab与ab的两个公共前缀（重复出现的子串）a、ab时，已经有了LCP(ab, aabaab) = 1个公共前缀（重复出现的子串）a已经计算过了，所以这时新的重复出现的子串的个数为LCP(abaab, ab) - LCP(ab, aabaab) = 1个
最后总结起来就是:
if(height[i] <= height[i - 1]) continue;
if(height[i] > height[i - 1] ) ans += height[i] - height[i - 1]
这就是最后的答案。
1 #include <map> 2 #include <set> 3 #include <stack> 4 #include <queue> 5 #include <cmath> 6 #include <ctime> 7 #include <vector> 8 #include <cstdio> 9 #include <cctype> 10 #include <cstring> 11 #include <cstdlib> 12 #include <iostream> 13 #include <algorithm> 14 using namespace std; 15 #define INF 0x3f3f3f3f 16 #define inf (-((LL)1<<40)) 17 #define lson k<<1, L, (L + R)>>1 18 #define rson k<<1|1, ((L + R)>>1) + 1, R 19 #define mem0(a) memset(a,0,sizeof(a)) 20 #define mem1(a) memset(a,-1,sizeof(a)) 21 #define mem(a, b) memset(a, b, sizeof(a)) 22 #define FIN freopen("in.txt", "r", stdin) 23 #define FOUT freopen("out.txt", "w", stdout) 24 #define rep(i, a, b) for(int i = a; i <= b; i ++) 25 #define dec(i, a, b) for(int i = a; i >= b; i --) 26 27 template<class T> T CMP_MIN(T a, T b) { return a < b; } 28 template<class T> T CMP_MAX(T a, T b) { return a > b; } 29 template<class T> T MAX(T a, T b) { return a > b ? a : b; } 30 template<class T> T MIN(T a, T b) { return a < b ? a : b; } 31 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; } 32 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b; } 33 34 //typedef __int64 LL; 35 typedef long long LL; 36 const int MAXN = 110000; 37 const int MAXM = 110000; 38 const double eps = 1e-4; 39 LL MOD = 1000000007; 40 41 struct SufArray { 42 char s[MAXN]; 43 int sa[MAXN], t[MAXN], t2[MAXN], c[MAXN], n, m; 44 int rnk[MAXN], height[MAXN]; 45 int mi[MAXN][20], idxK[MAXN]; 46 47 void init() { 48 mem0(s); 49 mem0(height); 50 } 51 void read_str() { 52 gets(s); 53 m = 128; 54 n = strlen(s); 55 s[n++] = ' '; 56 } 57 void build_sa() { 58 int *x = t, *y = t2; 59 rep (i, 0, m - 1) c[i] = 0; 60 rep (i, 0, n - 1) c[x[i] = s[i]] ++; 61 rep (i, 1, m - 1) c[i] += c[i - 1]; 62 dec (i, n - 1, 0) sa[--c[x[i]]] = i; 63 for(int k = 1; k <= n; k <<= 1) { 64 int p = 0; 65 rep (i, n - k, n - 1) y[p++] = i; 66 rep (i, 0, n - 1) if(sa[i] >= k) y[p++] = sa[i] - k; 67 rep (i, 0, m - 1) c[i] = 0; 68 rep (i, 0, n - 1) c[x[y[i]]] ++; 69 rep (i, 0, m - 1) c[i] += c[i - 1]; 70 dec (i, n - 1, 0) sa[--c[x[y[i]]]] = y[i]; 71 swap(x, y); 72 p = 1; 73 x[sa[0]] = 0; 74 rep (i, 1, n - 1) { 75 x[sa[i]] = y[sa[i - 1]] == y[sa[i]] && y[sa[i - 1] + k] == y[sa[i] + k] ? p - 1 : p++; 76 } 77 if(p >= n) break; 78 m = p; 79 } 80 } 81 void get_height() { 82 int k = 0; 83 rep (i, 0, n - 1) rnk[sa[i]] = i; 84 rep (i, 0, n - 1) { 85 if(k) k --; 86 int j = sa[rnk[i] - 1]; 87 while(s[i + k] == s[j + k]) k ++; 88 height[rnk[i]] = k; 89 } 90 } 91 void rmq_init(int *a, int n) { 92 rep (i, 0, n - 1) mi[i][0] = a[i]; 93 for(int j = 1; (1 << j) <= n; j ++) { 94 for(int i = 0; i + (1<<j) - 1 < n; i ++) { 95 mi[i][j] = min(mi[i][j - 1], mi[i + (1 << (j - 1))][j - 1]); 96 } 97 } 98 rep (len, 1, n) { 99 idxK[len] = 0; 100 while((1 << (idxK[len] + 1)) <= len) idxK[len] ++; 101 } 102 } 103 int rmq_min(int l, int r) { 104 int len = r - l + 1, k = idxK[len]; 105 return min(mi[l][k], mi[r - (1 << k) + 1][k]); 106 } 107 void lcp_init() { 108 get_height(); 109 rmq_init(height, n); 110 } 111 int get_lcp(int a, int b) { 112 if(a == b) return n - a - 1; 113 return rmq_min(min(rnk[a], rnk[b]) + 1, max(rnk[a], rnk[b])); 114 } 115 void solve() { 116 get_height(); 117 LL ans = 0, pre = 0; 118 rep (i, 1, n - 1) { 119 if(height[i] > pre) ans += height[i] - pre; 120 pre = height[i]; 121 } 122 cout << ans << endl; 123 } 124 }; 125 126 int T; 127 SufArray sa; 128 129 int main() 130 { 131 while(~scanf("%d%*c", &T)) while(T--){ 132 sa.init(); 133 sa.read_str(); 134 sa.build_sa(); 135 sa.solve(); 136 } 137 return 0; 138 } 139 /************************************************************** 140 Problem: 1632 141 User: csust_Rush 142 Language: C++ 143 Result: Accepted 144 Time:880 ms 145 Memory:13192 kb 146 ****************************************************************/