[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 }
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