解題區/解題區 - UVa

UVa 1186 - Chat Rooms

contents

  1. 1. Problem
  2. 2. Sample Input
  3. 3. Sample Output
  4. 4. Solution

Problem

聊天室會過濾過機器人般的垃圾留言，規則如下

  • 使用的單詞，出現連續子音的個數大於 5 個
  • 使用者最後送出的 10 個訊息中，有 2 行連續子音的個數大於 4 個，並且當前行也存在連續子音的個數大於 4 個。
  • 使用者最後送出的 10 個訊息中，與當前訊息相同的行數大於 1 個。

只要符合其中一條，訊息將無法發送，但是系統仍然會記錄下你曾經發送過的訊息，其他使用者無法看到被過濾的訊息。

Sample Input

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hello
how r u?
where r u from?
kjhh kh kgkjhg jhg
where r u from?
i am from London, Ontario, Canada
how r you nxw?
now
where r u from?
kjhh kh kgkjhg jhg
very good
it is very cold here.

Sample Output

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y
y
y
n
y
y
y
y
n
n
y
y

Solution

這題算簡單，但是題目描述過於鬼畜。

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#include <stdio.h> 
#include <vector>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <ctype.h>
using namespace std;
vector<string> toToken(string s) {
    stringstream sin(s);
    string token;
    vector<string> r;
    while (sin >> token)
        r.push_back(token);
    return r;
}
int isConsonant(char c) {
    c = tolower(c);
    if (c == 'a')	return 0;
    if (c == 'e')	return 0;
    if (c == 'i')	return 0;
    if (c == 'o')	return 0;
    if (c == 'u')	return 0;
    if (c == 'y')	return 0;
    return 1;
} 
int checkCond1(vector<string> &t) {
    for (int i = 0; i < t.size(); i++) {
        int c = 0;
        for (int j = 0; j < t[i].length(); j++) {
            if (isConsonant(t[i][j]))
                c++;
            else
                c = 0;
            if (c > 5)	return 0;
        }
    }
    return 1;
}
int checkCond2(vector<string> &t, vector<int> &D2) {
    int A = 0, B = 0;
    for (int i = 0; i < t.size(); i++) {
        int c = 0;
        for (int j = 0; j < t[i].length(); j++) {
            if (isConsonant(t[i][j]))
                c++;
            else
                c = 0;
            if (c > 4)	A = 1;
        }
    }
    for (int i = D2.size() - 1; i >= max(0, (int) D2.size() - 10); i--) {
        if (D2[i] > 0)
            B++;
    }
    return !(A && B > 2);
}
int checkCond3(string s, vector<string> &D3) {
    int A = 0;
    for (int i = D3.size() - 1; i >= max(0, (int) D3.size() - 10); i--) {
        if (D3[i] == s)
            A++;
        if (A > 1)
            return 0;
    }
    return 1;
}
int count4(vector<string> &t) {
    int A = 0;
    for (int i = 0; i < t.size(); i++) {
        int c = 0;
        for (int j = 0; j < t[i].length(); j++) {
            if (isConsonant(t[i][j]))
                c++;
            else
                c = 0;
            if (c > 4)	A = 1;
        }
    }
    return A;
}
int main() {
    int n;
    char s[256];
    while (scanf("%d", &n) == 1) {
        while (getchar() != '\n');
        vector< vector<string> > D;
        vector<int> D2;
        vector<string> D3;
        for (int i = 0; i < n; i++) { 
            gets(s);
            vector<string> t = toToken(s);
            int ok = 1;
            if (!checkCond1(t))
                ok = 0;
            else if (!checkCond2(t, D2))
                ok = 0;
            else if (!checkCond3(s, D3)) 
                ok = 0;
            D.push_back(t), D2.push_back(count4(t)), D3.push_back(s);
            puts(ok ? "y" : "n");
        } 
    }
    return 0;
}