链接：
codeforces.com/contest/432/problem/D
题意：
给你一个字符串，求每一个和前缀匹配的后缀在这个字符串中出现的次数
题解：
先算出lcp，找到sa[i]==0的位置标记为beg，和前缀匹配的后缀一定会出现beg的左边，这个想一想明白了
我们先算出beg左边每一个后缀和beg匹配的长度，beg右边的先放到一个vector中，便于之后二分查找
我们从beg向左遍历，对于每一个dp[i]，如果dp[i]==n-sa[i]，就可以更新结果了，从i到beg中间的都满足，同时还要加上beg右边的
代码：
31 int n, k; 32 int Rank[MAXN], tmp[MAXN]; 33 int sa[MAXN], lcp[MAXN]; 34 35 bool compare_sa(int i, int j) { 36 if (Rank[i] != Rank[j]) return Rank[i] < Rank[j]; 37 else { 38 int ri = i + k <= n ? Rank[i + k] : -1; 39 int rj = j + k <= n ? Rank[j + k] : -1; 40 return ri < rj; 41 } 42 } 43 44 void construct_sa(string S, int *sa) { 45 n = S.length(); 46 for (int i = 0; i <= n; i++) { 47 sa[i] = i; 48 Rank[i] = i < n ? S[i] : -1; 49 } 50 for (k = 1; k <= n; k *= 2) { 51 sort(sa, sa + n + 1, compare_sa); 52 tmp[sa[0]] = 0; 53 for (int i = 1; i <= n; i++) 54 tmp[sa[i]] = tmp[sa[i - 1]] + (compare_sa(sa[i - 1], sa[i]) ? 1 : 0); 55 for (int i = 0; i <= n; i++) Rank[i] = tmp[i]; 56 } 57 } 58 59 void construct_lcp(string S, int *sa, int *lcp) { 60 int n = S.length(); 61 for (int i = 0; i <= n; i++) Rank[sa[i]] = i; 62 int h = 0; 63 lcp[0] = 0; 64 for (int i = 0; i < n; i++) { 65 int j = sa[Rank[i] - 1]; 66 if (h > 0) h--; 67 for (; j + h < n && i + h < n; h++) 68 if (S[j + h] != S[i + h]) break; 69 lcp[Rank[i] - 1] = h; 70 } 71 } 72 73 int dp[MAXN]; 74 75 int main() { 76 ios::sync_with_stdio(false), cin.tie(0); 77 string s; 78 cin >> s; 79 construct_sa(s, sa); 80 construct_lcp(s, sa, lcp); 81 int beg; 82 rep(i, 0, n + 1) if (sa[i] == 0) { 83 beg = i; 84 break; 85 } 86 memset(dp, 0x3f, sizeof(dp)); 87 dp[beg] = INF; 88 per(i, 0, beg) dp[i] = min(dp[i + 1], lcp[i]); 89 rep(i, beg + 1, n + 1) dp[i] = min(dp[i - 1], lcp[i - 1]); 90 VI v; 91 rep(i, beg + 1, n + 1) if (dp[i]) v.pb(dp[i]); 92 sort(all(v)); 93 map<int, int> mmp; 94 per(i, 0, beg) { 95 if (!dp[i]) break; 96 if (dp[i] == n - sa[i]) { 97 mmp[dp[i]] += beg - i + 1; 98 mmp[dp[i]] += v.end() - lower_bound(all(v), dp[i]); 99 } 100 } 101 cout << mmp.size() + 1 << endl; 102 for (auto it : mmp) cout << it.first << ' ' << it.second << endl; 103 cout << n << ' ' << 1 << endl; 104 return 0; 105 }