Problem Description

String matching is a common type of problem in computer science. One string matching problem is as following:

Given a string s[0…len?1], please calculate the length of the longest common prefix of s[i…len?1] and s[0…len?1] for each i>0.

I believe everyone can do it by brute force.
The pseudo code of the brute force approach is as the following:

We are wondering, for any given string, what is the number of compare operations invoked if we use the above algorithm. Please tell us the answer before we attempt to run this algorithm.

Input

The first line contains an integer T, denoting the number of test cases.
Each test case contains one string in a line consisting of printable ASCII characters except space.

* 1≤T≤30

* string length ≤106 for every string

Output

For each test, print an integer in one line indicating the number of compare operations invoked if we run the algorithm in the statement against the input string.

Sample Input

3
_Happy_New_Year_
ywwyww
zjczzzjczjczzzjc

Sample Output

17
7
32

題意：t 組數(shù)據(jù)，每組給出一個字符串 S，現(xiàn)在有一個函數(shù)，每次會暴力的跑?S 的所有后綴與字符串 S 的最長公共前綴，問這個函數(shù)會運(yùn)行多少次

思路：

實(shí)質(zhì)是求?S 的每個后綴與 S 的最長公共前綴的比較次數(shù)之和

擴(kuò)展 KMP 模版題，求出 extend 數(shù)組后比較累加即可

Source Program

#include<iostream> #include<cstdio> #include<cstdlib> #include<string> #include<cstring> #include<cmath> #include<ctime> #include<algorithm> #include<utility> #include<stack> #include<queue> #include<vector> #include<set> #include<map> #include<unordered_map> #include<bitset> #define PI acos(-1.0) #define INF 0x3f3f3f3f #define LL long long #define Pair pair<int,int> LL quickPow(LL a,LL b){ LL res=1; while(b){if(b&1)res*=a; a*=a; b>>=1;} return res; } LL multMod(LL a,LL b,LL mod){ a%=mod; b%=mod; LL res=0; while(b){if(b&1)res=(res+a)%mod; a=(a<<=1)%mod; b>>=1; } return res%mod;} LL quickPowMod(LL a, LL b,LL mod){ LL res=1,k=a; while(b){if((b&1))res=multMod(res,k,mod)%mod; k=multMod(k,k,mod)%mod; b>>=1;} return res%mod;} LL getInv(LL a,LL mod){ return quickPowMod(a,mod-2,mod); } LL GCD(LL x,LL y){ return !y?x:GCD(y,x%y); } LL LCM(LL x,LL y){ return x/GCD(x,y)*y; } const double EPS = 1E-10; const int MOD = 998244353; const int N = 1000000+5; const int dx[] = {-1,1,0,0,1,-1,1,1}; const int dy[] = {0,0,-1,1,-1,1,-1,1}; using namespace std;int Next[N]; int extend[N]; void getNext(char *str) {int len = strlen(str);Next[0] = len;int i = 0;while (str[i] == str[i + 1] && i + 1 < len)i++;Next[1] = i;int pos = 1;for (i = 2; i < len; i++) {if (Next[i - pos] + i < Next[pos] + pos)Next[i] = Next[i - pos];else {int j = Next[pos] + pos - i;if (j < 0)j = 0;while (i + j < len && str[j] == str[j + i])j++;Next[i] = j;pos = i;}} } void exKMP(char *s1, char *s2) {getNext(s2);int len1 = strlen(s1);int len2 = strlen(s2);int i = 0;while (s1[i] == s2[i] && i < len2 && i < len1)i++;extend[0] = i;int pos = 0;for (i = 1; i < len1; i++) {if (Next[i - pos] + i < extend[pos] + pos)extend[i] = Next[i - pos];else {int j = extend[pos] + pos - i;if (j < 0)j = 0;while (i + j < len1 && j < len2 && s1[j + i] == s2[j])j++;extend[i] = j;pos = i;}} }char s[N]; int main(){int t;scanf("%d",&t);while(t--){memset(Next,0,sizeof(Next));memset(extend,0,sizeof(extend));scanf("%s",s);exKMP(s,s);LL res=0;int len=strlen(s);for(int i=1;i<len;i++){res+=extend[i]+1;if(extend[i]+i>=len)res--;}printf("%lld\n",res);}return 0; }

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