[hdu3336]kmp(后缀数组)

题意：求字符串s的所有前缀出现次数之和。

http://www.cnblogs.com/jklongint/p/4446117.html

思路：用kmp做，简单且效率高。以前缀结尾的位置分类，令dp[i]为以结尾位置在i的前缀数量，那么dp[i] = cnt(j)(j~i是前缀)，而由kmp的next函数的转移性质，可得如下递推方程:dp[i] = dp[next[i]] + 1，把这个递推式不断展开，也就是i = next[i]不断迭代，那么+1的个数就是dp[i] = cnt(j)(j~i是前缀)这个里面的cnt(j)了。这就是next数组的性质，实在巧妙！

  1 #pragma comment(linker, "/STACK:10240000,10240000")
  2 
  3 #include <iostream>
  4 #include <cstdio>
  5 #include <algorithm>
  6 #include <cstdlib>
  7 #include <cstring>
  8 #include <map>
  9 #include <queue>
 10 #include <deque>
 11 #include <cmath>
 12 #include <vector>
 13 #include <ctime>
 14 #include <cctype>
 15 #include <set>
 16 #include <bitset>
 17 #include <functional>
 18 #include <numeric>
 19 #include <stdexcept>
 20 #include <utility>
 21 
 22 using namespace std;
 23 
 24 #define mem0(a) memset(a, 0, sizeof(a))
 25 #define mem_1(a) memset(a, -1, sizeof(a))
 26 #define lson l, m, rt << 1
 27 #define rson m + 1, r, rt << 1 | 1
 28 #define define_m int m = (l + r) >> 1
 29 #define rep_up0(a, b) for (int a = 0; a < (b); a++)
 30 #define rep_up1(a, b) for (int a = 1; a <= (b); a++)
 31 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
 32 #define rep_down1(a, b) for (int a = b; a > 0; a--)
 33 #define all(a) (a).begin(), (a).end()
 34 #define lowbit(x) ((x) & (-(x)))
 35 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
 36 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
 37 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
 38 #define pchr(a) putchar(a)
 39 #define pstr(a) printf("%s", a)
 40 #define sstr(a) scanf("%s", a)
 41 #define sint(a) scanf("%d", &a)
 42 #define sint2(a, b) scanf("%d%d", &a, &b)
 43 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
 44 #define pint(a) printf("%d\n", a)
 45 #define test_print1(a) cout << "var1 = " << a << endl
 46 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
 47 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl
 48 
 49 typedef double db;
 50 typedef long long LL;
 51 typedef pair<int, int> pii;
 52 typedef multiset<int> msi;
 53 typedef set<int> si;
 54 typedef vector<int> vi;
 55 typedef map<int, int> mii;
 56 
 57 const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1};
 58 const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 };
 59 const int maxn = 2e5 + 7;
 60 const int md = 10007;
 61 const int inf = 1e9 + 7;
 62 const LL inf_L = 1e18 + 7;
 63 const double pi = acos(-1.0);
 64 const double eps = 1e-6;
 65 
 66 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}
 67 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
 68 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
 69 template<class T>T condition(bool f, T a, T b){return f?a:b;}
 70 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
 71 int make_id(int x, int y, int n) { return x * n + y; }
 72 
 73 struct KMP {
 74     int next[1000010];
 75     void GetNext(char s[]) {
 76         mem0(next);
 77         next[0] = next[1] = 0;
 78         for(int i = 1; s[i]; i++) {
 79             int j = next[i];
 80             while(j && s[i] != s[j]) j = next[j];
 81             next[i + 1] = s[j] == s[i]? j + 1 : 0;
 82         }
 83     }
 84 };
 85 
 86 KMP kmp;
 87 int dp[maxn];
 88 char s[maxn];
 89 
 90 int main() {
 91     //freopen("in.txt", "r", stdin);
 92     int T, n;
 93     cin >> T;
 94     while (T--) {
 95         scanf("%d%s", &n, s);
 96         kmp.GetNext(s);
 97         mem0(dp);
 98         int ans = n;
 99         rep_up1(i, n) {
100             if (kmp.next[i] == 0) continue;
101             dp[i] = dp[kmp.next[i]] + 1;
102             dp[i] %= md;
103             ans += dp[i];
104             ans %= md;
105         }
106         cout << ans << endl;
107     }
108     return 0;
109 }

View Code

 

转载于:https://www.cnblogs.com/jklongint/p/4449244.html