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hdu 5510 Bazinga(字符串kmp)
Bazinga
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2287 Accepted Submission(s): 713
Problem Description
Ladies and gentlemen, please sit up straight.
Don‘t tilt your head. I‘m serious.
For n given strings S1,S2,?,Sn, labelled from 1 to n, you should find the largest i (1≤i≤n) such that there exists an integer j (1≤j<i) and Sj is not a substring of Si.
A substring of a string Si is another string that occurs in Si. For example, ``ruiz" is a substring of ``ruizhang", and ``rzhang" is not a substring of ``ruizhang".
Don‘t tilt your head. I‘m serious.
For n given strings S1,S2,?,Sn, labelled from 1 to n, you should find the largest i (1≤i≤n) such that there exists an integer j (1≤j<i) and Sj is not a substring of Si.
A substring of a string Si is another string that occurs in Si. For example, ``ruiz" is a substring of ``ruizhang", and ``rzhang" is not a substring of ``ruizhang".
Input
The first line contains an integer t (1≤t≤50) which is the number of test cases.
For each test case, the first line is the positive integer n (1≤n≤500) and in the following n lines list are the strings S1,S2,?,Sn.
All strings are given in lower-case letters and strings are no longer than 2000 letters.
For each test case, the first line is the positive integer n (1≤n≤500) and in the following n lines list are the strings S1,S2,?,Sn.
All strings are given in lower-case letters and strings are no longer than 2000 letters.
Output
For each test case, output the largest label you get. If it does not exist, output −1.
Sample Input
45ababczabcabcdzabcd4youlovinyouaboutlovinyouallaboutlovinyou5dedefabcdabcdeabcdef3abaccc
Sample Output
Case #1: 4Case #2: -1Case #3: 4Case #4: 3
/*hdu 5510 Bazinga(字符串kmp)problem:给你n个字符串,求一个最大的序号i. 使1~i-1的字符串中有一个不是str[i]的子串solve:最开始看见想的是字典树.但是感觉很麻烦,目测会超时.然后想的是kmp,但是优化有问题. 每次倒着往前搜,想的是如果1~i-1的所有都是i的子串. 那么搜索到i时就可以直接返回true.如果不匹配就退出,否则就一直搜索下去. 结果TLE了....很悲伤后来发现正确优化:如果i是k的子串. 那么匹配的时候就不需要匹配i,因为如果当前与k匹配,那么就一定与i匹配....所以过程中标记一下不用匹配的就行了hhh-2016-08-29 21:00:19*/#pragma comment(linker,"/STACK:124000000,124000000")#include <algorithm>#include <iostream>#include <cstdlib>#include <cstdio>#include <cstring>#include <vector>#include <math.h>#include <queue>#include <set>#include <map>#define lson i<<1#define rson i<<1|1#define ll long long#define clr(a,b) memset(a,b,sizeof(a))#define scanfi(a) scanf("%d",&a)#define scanfs(a) scanf("%s",a)#define scanfl(a) scanf("%I64d",&a)#define key_val ch[ch[root][1]][0]#define inf 1e9using namespace std;const ll mod = 1e9+7;const int maxn = 2005;char str[505][2004];int flag[maxn];void pre_kmp(char x[],int m,int kmpnext[]){ int i,j; j = kmpnext[0] = -1; i = 0; while(i < m) { while(j != -1 && x[i] != x[j]) j = kmpnext[j]; if(x[++i] == x[++j]) kmpnext[i] = kmpnext[j]; else kmpnext[i] = j; }}int nex[2005];int kmp(char x[],char y[]){ int m = strlen(x); int n = strlen(y); int i,j; clr(nex,0); pre_kmp(x,m,nex); i = j = 0; while(i < n) { while(j != -1 && y[i] != x[j]) j = nex[j]; i++,j++; if(j >= m) { return true; } } return false;}int main(){ int T,n;// freopen("in.txt","r",stdin); scanfi(T); int cas = 1; while(T--) { clr(flag,0); scanfi(n); int ans = -1; for(int i = 1; i <= n; i++) { scanfs(str[i]); for(int j =1;j < i;j++) { if(flag[j]) continue; if(kmp(str[j],str[i])) flag[j] = 1; else { ans = i; } } } printf("Case #%d: %d\n",cas++,ans); } return 0;}
hdu 5510 Bazinga(字符串kmp)
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