hdu 5918 Sequence I (kmp)
2016-10-04 20:10
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Sequence I
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Problem Description
Mr. Frog has two sequences a1,a2,⋯,an and b1,b2,⋯,bm and a number p. He wants to know the number of positions q such that sequence b1,b2,⋯,bm is exactly the sequence aq,aq+p,aq+2p,⋯,aq+(m−1)p where q+(m−1)p≤n and q≥1.
Input
The first line contains only one integer T≤100, which indicates the number of test cases.
Each test case contains three lines.
The first line contains three space-separated integers 1≤n≤106,1≤m≤106 and 1≤p≤106.
The second line contains n integers a1,a2,⋯,an(1≤ai≤109).
the third line contains m integers b1,b2,⋯,bm(1≤bi≤109).
Output
For each test case, output one line “Case #x: y”, where x is the case number (starting from 1) and y is the number of valid q’s.
Sample Input
2
6 3 1
1 2 3 1 2 3
1 2 3
6 3 2
1 3 2 2 3 1
1 2 3
Sample Output
Case #1: 2
Case #2: 1
这题本来是觉得似乎不能用kmp,结果重现赛所有的CE都是kmp。然后我就开始方了。
然后开始思考,kmp模板里面的j = nxt[m]这个部分如果直接套用肯定会跳过一些可能的匹配导致出错,但只考虑和模式串的匹配的话,kmp还是有用武之地的。
于是把kmp胡改一通,枚举所有匹配起点,每次匹配完一个模式串长度,原串匹配下标加一,模式串下标变0重新开始。
本以为会T,结果并没有。对kmp的复杂度一直都不是很懂,改变之前最大复杂度是O(n + m)这个是可以想清楚的,但是改变之后复杂度应该是会相应增大一些,不是很清楚增大的幅度,结果写的时候没法好好估计时间,感觉会是个很浪费时间的点。
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <map> #include <cmath> #define MAX 1000010 #define M 15 #define INF 0x3f3f3f3f #define PI acos(-1.0) #define eps 1e-8 using namespace std; int t, n, m, p, ans; int txt[MAX], nxt[MAX], T[MAX]; void getnxt()//模板 { nxt[0] = -1; int i = 0, j = -1; while(i < m) { while(j >= 0 && txt[i] != txt[j]) j = nxt[j]; j++; i++; if(txt[i] == txt[j]) nxt[i] = nxt[j]; else nxt[i] = j; } } void kmp(int p) { getnxt(); int i = 0, j = 0, cnt = 0; while(i < n) { while(j >= 0 && T[i] != txt[j]) j = nxt[j]; i += p; cnt++;//与模板不同的部分 j++; if(j == m)//与模板不同的部分 { ans++; j = 0; i -= m * p; i++; cnt = 0; } else if(cnt >= m)//比模板增加的部分 { i -= m * p; i++; j = 0; cnt = 0; } } } int main() { scanf("%d", &t); for(int i = 1; i <= t; i++) { ans = 0; scanf("%d %d %d", &n, &m, &p); for(int j = 0; j < n; j++) scanf("%d", &T[j]); for(int j = 0; j < m; j++) scanf("%d", &txt[j]); kmp(p); printf("Case #%d: %d\n", i, ans); } return 0; }
运行结果:
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