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題目傳送門

1 /* 2 題意：求出一個串的前綴與后綴相同的字串的長度 3 KMP：nex[]就有這樣的性質(zhì)，倒過來輸出就行了 4 */ 5 /************************************************ 6 * Author :Running_Time 7 * Created Time :2015-8-10 11:29:25 8 * File Name :POJ_2752.cpp 9 ************************************************/ 10 11 #include <cstdio> 12 #include <algorithm> 13 #include <iostream> 14 #include <sstream> 15 #include <cstring> 16 #include <cmath> 17 #include <string> 18 #include <vector> 19 #include <queue> 20 #include <deque> 21 #include <stack> 22 #include <list> 23 #include <map> 24 #include <set> 25 #include <bitset> 26 #include <cstdlib> 27 #include <ctime> 28 using namespace std; 29 30 #define lson l, mid, rt << 1 31 #define rson mid + 1, r, rt << 1 | 1 32 typedef long long ll; 33 const int MAXN = 4e5 + 10; 34 const int INF = 0x3f3f3f3f; 35 const int MOD = 1e9 + 7; 36 int nex[MAXN]; 37 char str[MAXN]; 38 39 void get_nex(void) { 40 int len = strlen (str); 41 int i = 0, j = -1; nex[0] = -1; 42 while (i < len) { 43 if (j == -1 || str[j] == str[i]) { 44 j++; i++; nex[i] = j; 45 } 46 else j = nex[j]; 47 } 48 vector<int> ans; 49 while (i != -1) { 50 ans.push_back (i); i = nex[i]; 51 } 52 for (int i=ans.size ()-2; i>=0; --i) { 53 printf ("%d ", ans[i]); 54 } 55 puts (""); 56 } 57 58 int main(void) { //POJ 2752 Seek the Name, Seek the Fame 59 while (scanf ("%s", str) == 1) { 60 get_nex (); 61 } 62 63 return 0; 64 }

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轉(zhuǎn)載于:https://www.cnblogs.com/Running-Time/p/4717781.html

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