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

---

题目

---

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

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 349  Solved: 168
[Submit][Status]

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".

---

题解

---

暴力就可以了。

---

代码

---

/*Author:WNJXYK*/
#include<cstdio>
#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
using namespace std;

#define LL long long
#define Inf 2147483647
#define InfL 10000000000LL

inline void swap(int &x,int &y){int tmp=x;x=y;y=tmp;}
inline void swap(LL &x,LL &y){LL tmp=x;x=y;y=tmp;}
inline int remin(int a,int b){if (a<b) return a;return b;}
inline int remax(int a,int b){if (a>b) return a;return b;}
inline LL remin(LL a,LL b){if (a<b) return a;return b;}
inline LL remax(LL a,LL b){if (a>b) return a;return b;}

string name[1001];
string power[101];
int ans[1001];
int n,m;
inline char upcase(char x){
	if ('a'<=x && x<='z') return x-'a'+'A';
	return x; 
}
inline int judge(int x,int y){
	int lx=power[x].length()-1,ly=name[y].length()-1;
	for (int i=0;i<=ly-lx;i++){
		bool flag=true;
		int now=i;
		if (upcase(name[y][i])==upcase(power[x][0]))
		for (int j=1;j<=lx;j++){
			now++;
			while(now<=ly && upcase(power[x][j])!=upcase(name[y][now])) now++;
			if (now>ly) {
				flag=false;
				break;
			}
		}
		else
		continue;
		if (flag) return 1;
	}
	return 0;
} 
int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++) cin>>name[i];
	for (int j=1;j<=m;j++) cin>>power[j];
	for (int i=1;i<=m;i++){
		for (int j=1;j<=n;j++){
			ans[j]+=judge(i,j);
		}
	}
	for (int i=1;i<=n;i++)printf("%d
",ans[i]);
	return 0;
}