BZOJ 1622: [Usaco2008 Open]Word Power 名字的能量 (暴力)

1622: [Usaco2008 Open]Word Power 名字的能量

Time Limit: 5 Sec  Memory Limit: 64 MB Submit: 462  Solved: 228 [Submit][Status][Discuss]

Description

    约翰想要计算他那N(1≤N≤1000)只奶牛的名字的能量.每只奶牛的名字由不超过1000个字待构成,没有一个名字是空字体串,  约翰有一张“能量字符串表”,上面有M(1≤M≤100)个代表能量的字符串.每个字符串由不超过30个字体构成,同样不存在空字符串.一个奶牛的名字蕴含多少个能量字符串,这个名字就有多少能量.所谓“蕴含”,是指某个能量字符串的所有字符都在名字串中按顺序出现(不一定一个紧接着一个).

    所有的大写字母和小写字母都是等价的.比如,在贝茜的名字“Bessie”里,蕴含有“Be”

“sI”“EE”以及“Es”等等字符串,但不蕴含“lS”或“eB”.请帮约翰计算他的奶牛的名字的能量.

Input

    第1行输入两个整数N和M,之后N行每行输入一个奶牛的名字,之后M行每行输入一个能量字符串.

Output

    一共N行,每行一个整数,依次表示一个名字的能量.

Sample Input

5 3 Bessie Jonathan Montgomery Alicia Angola se nGo Ont

INPUT DETAILS:

There are 5 cows, and their names are "Bessie", "Jonathan", "Montgomery", "Alicia", and "Angola". The 3 good strings are "se", "nGo", and "Ont".

Sample Output

1 1 2 0 1

OUTPUT DETAILS:

"Bessie" contains "se", "Jonathan" contains "Ont", "Montgomery" contains both "nGo" and "Ont", Alicia contains none of the good strings, and "Angola" contains "nGo".

思路:复杂度1E8..5S的时限。。。暴力。。

/* ***********************************************
Author :111qqz
Created Time :2016年04月04日 星期一 02时10分20秒
File Name :code/bzoj/1622.cpp
************************************************ */
 1#include <cstdio>
 2#include <cstring>
 3#include <iostream>
 4#include <algorithm>
 5#include <vector>
 6#include <queue>
 7#include <set>
 8#include <map>
 9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define fst first
14#define sec second
15#define lson l,m,rt<<1
16#define rson m+1,r,rt<<1|1
17#define ms(a,x) memset(a,x,sizeof(a))
18typedef long long LL;
19#define pi pair < int ,int >
20#define MP make_pair
1using namespace std;
2const double eps = 1E-8;
3const int dx4[4]={1,0,0,-1};
4const int dy4[4]={0,-1,1,0};
5const int inf = 0x3f3f3f3f;
6const int N=1E3+7;
7string cow[N];
8string eng[105];
9int n,m;
1string  aha ( string x)
2{
3    int len = x.length();
4    for ( int i = 0 ; i < len ; i++)
5    {
6	if (x[i]>='A'&&x[i]<='Z') x[i] = char(x[i]+32);
7    }
8    return x;
9}
 1bool ok (string x,string y)
 2{
 3  //  cout<<"x:"<<x<<" y:"<<y<<endl;
 4    int lx = x.length();
 5    int ly = y.length();
 6    int j = 0;
 7    for ( int i = 0 ; i < lx ; i++)
 8    {
 9	if (x[i]==y[j]) j++;
10	if (j>=ly) return true;
11    }
12    return false;
1}
2int main()
3{
4	#ifndef  ONLINE_JUDGE 
5	freopen("code/in.txt","r",stdin);
6  #endif
1	ios::sync_with_stdio(false);
2	cin>>n>>m;
3	for ( int i = 1 ; i <= n ; i++) cin>>cow[i],cow[i]=aha(cow[i]);
4	for ( int i = 1 ; i <= m ; i++) cin>>eng[i],eng[i]=aha(eng[i]);
	for ( int i = 1 ; i <= n ; i++)
	{
1	    int ans =  0;
2	    for ( int j = 1 ; j <= m ; j++)
3	    {
4		if (ok(cow[i],eng[j])) ans++;
5	    }
6	    cout<<ans<<endl;
7	}
1  #ifndef ONLINE_JUDGE  
2  fclose(stdin);
3  #endif
4    return 0;
5}