Excuses, Excuses!
Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld
& %llu
Description
| Excuses, Excuses! |
Judge Ito is having a problem with people subpoenaed for jury duty giving rather lame excuses in order to avoid serving. In order to reduce the amount of time required listening to goofy excuses, Judge Ito has asked that you write a program that will search
for a list of keywords in a list of excuses identifying lame excuses. Keywords can be matched in an excuse regardless of case.
Input
Input to your program will consist of multiple sets of data.
- Line 1 of each set will contain exactly two integers. The first number (
) defines the number of keywords to be used in the
search. The second number (
) defines the number of excuses in the set to be searched. - Lines 2 through K+1 each contain exactly one keyword.
- Lines K+2 through K+1+E each contain exactly one excuse.
- All keywords in the keyword list will contain only contiguous lower case alphabetic characters of length L (
) and
will occupy columns 1 through L in the input line. - All excuses can contain any upper or lower case alphanumeric character, a space, or any of the following punctuation marks [
SPMamp".,!?&] not including the square brackets and will not exceed 70 characters in length. - Excuses will contain at least 1 non-space character.
) defines the number of keywords to be used in the
) defines the number of excuses in the set to be searched.
) andOutput
For each input set, you are to print the worst excuse(s) from the list.
- The worst excuse(s) is/are defined as the excuse(s) which contains the largest number of incidences of keywords.
- If a keyword occurs more than once in an excuse, each occurrance is considered a separate incidence.
- A keyword ``occurs" in an excuse if and only if it exists in the string in contiguous form and is delimited by the beginning or end of the line or any non-alphabetic character or a space.
For each set of input, you are to print a single line with the number of the set immediately after the string ``Excuse Set #". (See the Sample Output). The following line(s) is/are to contain the worst excuse(s) one per line exactly as read in. If
there is more than one worst excuse, you may print them in any order.
After each set of output, you should print a blank line.
Sample Input
5 3 dog ate homework canary died My dog ate my homework. Can you believe my dog died after eating my canary... AND MY HOMEWORK? This excuse is so good that it contain 0 keywords. 6 5 superhighway crazy thermonuclear bedroom war building I am having a superhighway built in my bedroom. I am actually crazy. 1234567890.....,,,,,0987654321?????!!!!!! There was a thermonuclear war! I ate my dog, my canary, and my homework ... note outdated keywords?
Sample Output
Excuse Set #1 Can you believe my dog died after eating my canary... AND MY HOMEWORK? Excuse Set #2 I am having a superhighway built in my bedroom. There was a thermonuclear war!
题意：前面给出n个特殊单词，后面给出m句话，问这m句话中特殊单词的个数最多的是哪句话？如果有两句话的特殊单词数相同按输入的顺序同时输出（因为不知道要按输入时的顺序输出，改了一天半，把所有可能犯的错全改了，就是交不上，真是醉了）
POJ的数据略水，还是说UVA的数据太硬...在UVA上WA的代码在POJ上直接Accepted。
UVA AC代码：
#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<string>
#include<map>
using namespace std;
int n,m;
struct node
{
int x,y;
} q[100010];
int main()
{
int e = 0;
while(scanf("%d%d",&n,&m)!=EOF)
{
char a[10001];
map<string,int>p;
map<string,int>::iterator it;
for(int i=0; i<n; i++)
{
scanf("%s",a);
for(int j=0; a[j]!=‘\0‘; j++)
{
if(a[j]>=‘A‘ && a[j]<=‘Z‘)
{
a[j] = a[j] + 32;
}
}
p[a] = 1;
}
char b[3001][101];
char str[3001][110];
memset(q,0,sizeof(q));
getchar();
for(int i=0; i<m; i++)
{
gets(b[i]);
int t = 0;
int k = 0;
for(int j=0; b[i][j]!=‘\0‘; j++)
{
if((b[i][j] >=‘a‘ && b[i][j]<=‘z‘) || (b[i][j]>=‘A‘ && b[i][j]<=‘Z‘))
{
str[t][k++] = b[i][j];
}
else
{
str[t][k] = ‘\0‘;
t++;
k = 0;
}
}
str[t][k] = ‘\0‘;
q[i].x = i;
for(int j=0; j<=t; j++)
{
if((str[j][0] >=‘A‘ && str[j][0]<=‘Z‘) || (str[j][0]>=‘a‘ && str[j][0]<=‘z‘))
{
for(int v=0; str[j][v]!=‘\0‘; v++)
{
if(str[j][v]>=‘A‘ && str[j][v]<=‘Z‘)
{
str[j][v] = str[j][v] + 32;
}
}
if((str[j][0] >=‘a‘ && str[j][0]<=‘z‘) && p[str[j]] == 1)
{
q[i].y++;
}
}
}
}
int max = 0;
for(int i=0;i<m;i++)
{
if(max<q[i].y)
{
max = q[i].y;
}
}
printf("Excuse Set #%d\n",++e);
for(int i=0;i<m;i++)
{
if(max == q[i].y)
{
printf("%s\n",b[q[i].x]);
}
}
printf("\n");
}
return 0;
}
UVA提交WA，POJ提交AC代码：
#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<string>
#include<map>
using namespace std;
int n,m;
struct node
{
int x,y;
} q[100010];
int cmp(const void *a,const void *b)
{
return (*(struct node*)a).y - (*(struct node*)b).y;
}
int main()
{
int e = 0;
while(scanf("%d%d",&n,&m)!=EOF)
{
char a[10001];
map<string,int>p;
map<string,int>::iterator it;
for(int i=0; i<n; i++)
{
scanf("%s",a);
for(int j=0; a[j]!=‘\0‘; j++)
{
if(a[j]>=‘A‘ && a[j]<=‘Z‘)
{
a[j] = a[j] + 32;
}
}
p[a] = 1;
}
char b[3001][101];
char str[3001][110];
memset(q,0,sizeof(q));
getchar();
for(int i=0; i<m; i++)
{
gets(b[i]);
int t = 0;
int k = 0;
for(int j=0; b[i][j]!=‘\0‘; j++)
{
if((b[i][j] >=‘a‘ && b[i][j]<=‘z‘) || (b[i][j]>=‘A‘ && b[i][j]<=‘Z‘))
{
str[t][k++] = b[i][j];
}
else
{
str[t][k] = ‘\0‘;
t++;
k = 0;
}
}
str[t][k] = ‘\0‘;
q[i].x = i;
for(int j=0; j<=t; j++)
{
if((str[j][0] >=‘A‘ && str[j][0]<=‘Z‘) || (str[j][0]>=‘a‘ && str[j][0]<=‘z‘))
{
for(int v=0; str[j][v]!=‘\0‘; v++)
{
if(str[j][v]>=‘A‘ && str[j][v]<=‘Z‘)
{
str[j][v] = str[j][v] + 32;
}
}
if((str[j][0] >=‘a‘ && str[j][0]<=‘z‘) && p[str[j]] == 1)
{
//printf("%s\n",str[j]);
q[i].y++;
}
}
}
}
qsort(q,m,sizeof(q[0]),cmp);
printf("Excuse Set #%d\n",++e);
printf("%s\n",b[q[m-1].x]);
for(int i=m-2; i>=0; i--)
{
if(q[i].y == q[i+1].y)
{
printf("%s\n",b[q[i].x]);
}
else
{
break;
}
}
printf("\n");
}
return 0;
}