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  • [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
    ", 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
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  • 原文地址:https://www.cnblogs.com/jklongint/p/4449244.html
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