hdu5510 Bazinga 暴力+剪枝

Bazinga

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3567    Accepted Submission(s): 1146

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".

 

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.

 

Output

For each test case, output the largest label you get. If it does not exist, output −1.

 

Sample Input

4 5 ab abc zabc abcd zabcd 4 you lovinyou aboutlovinyou allaboutlovinyou 5 de def abcd abcde abcdef 3 a ba ccc

 

Sample Output

Case #1: 4 Case #2: -1 Case #3: 4 Case #4: 3

 

i從0到n - 1遍歷，用一個vis標記j是否是其後面某個i的子串，是的話vis[j] = 1，否則vis[j] = 0，在遍歷i = i + 1的時候

如果遍歷前面的vis[j] == 1的話，就不算這個j，原因是，從i - 1到j + 1這一段，假如遇到了一個x不是i的子串，那麼直接記錄了flag = i，繼續更新前面vis[x] == 0的看是不是i的子串，是的話修改vis[x] = 1；如果是i的子串的話，那麼前面打上vis[x] == 1的肯定是後面的子串，因爲我們是從後往前掃的，在掃到後面的時候，如果後面的是i的子串了的話，那麼前面的還是這個子串的子串，肯定是i的子串，沒有必要匹配了，達到剪枝的目的

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

int T, n, kase;
char a[505][2005];
int vis[505];

int main()
{
	cin >> T;
	kase = 0;
	while (T--) {
		scanf("%d", &n);
		memset(vis, 0, sizeof(vis));
		for (int i = 0; i < n; i++) {
			scanf("%s", a[i]);
		}
		int flag = 0;
		for (int i = 0; i < n; i++) {
			for (int j = i - 1; j >= 0; j--) {
				if (!vis[j]) {
					if (strstr(a[i], a[j]) == NULL) flag = i + 1;
					else vis[j] = 1;
				}
			}
		}
		if (flag == 0) flag = -1;
		printf("Case #%d: %d\n", ++kase, flag);
	}
	return 0;
}