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  • A. Vasya and Football

    A. Vasya and Football
    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Vasya has started watching football games. He has learned that for some fouls the players receive yellow cards, and for some fouls they receive red cards. A player who receives the second yellow card automatically receives a red card.

    Vasya is watching a recorded football match now and makes notes of all the fouls that he would give a card for. Help Vasya determine all the moments in time when players would be given red cards if Vasya were the judge. For each player, Vasya wants to know only thefirst moment of time when he would receive a red card from Vasya.

    Input

    The first line contains the name of the team playing at home. The second line contains the name of the team playing away. Both lines are not empty. The lengths of both lines do not exceed 20. Each line contains only of large English letters. The names of the teams are distinct.

    Next follows number n (1 ≤ n ≤ 90) — the number of fouls.

    Each of the following n lines contains information about a foul in the following form:

    • first goes number t (1 ≤ t ≤ 90) — the minute when the foul occurs;
    • then goes letter "h" or letter "a" — if the letter is "h", then the card was given to a home team player, otherwise the card was given to an away team player;
    • then goes the player's number m (1 ≤ m ≤ 99);
    • then goes letter "y" or letter "r" — if the letter is "y", that means that the yellow card was given, otherwise the red card was given.

    The players from different teams can have the same number. The players within one team have distinct numbers. The fouls go chronologically, no two fouls happened at the same minute.

    Output

    For each event when a player received his first red card in a chronological order print a string containing the following information:

    • The name of the team to which the player belongs;
    • the player's number in his team;
    • the minute when he received the card.

    If no player received a card, then you do not need to print anything.

    It is possible case that the program will not print anything to the output (if there were no red cards).

    Sample test(s)
    input
    MC
    CSKA
    9
    28 a 3 y
    62 h 25 y
    66 h 42 y
    70 h 25 y
    77 a 4 y
    79 a 25 y
    82 h 42 r
    89 h 16 y
    90 a 13 r
    
    output
    MC 25 70
    MC 42 82
    CSKA 13 90
    #include <iostream>
    #include <cstring>
    #include <string>
    #include <cstdio>
    #include <cstdlib>
    #include <algorithm>
    using namespace std;
    
    int b[100], c1[110], c2[110], c3[110], c4[110];
    
    struct node {
    	int t;
    	char c;
    	int m;
    	char d;
    }a[100];
    
    int cmp(node a, node b) {
    	return a.t < b.t;
    }
    
    int main() {
    	string str1, str2;
    	cin >> str1 >> str2;
    	int n;
    	cin >> n;
    	memset(a, 0, sizeof(a));
    
    	for (int i = 0; i < n; i++)
    		cin >> a[i].t >> a[i].c >> a[i].m >> a[i].d;
    	sort(a, a + n, cmp);
    	
    	memset(b, 0, sizeof(b));
    	memset(c1, 0, sizeof(c1));
    	memset(c2, 0, sizeof(c2));
    	memset(c3, 0, sizeof(c2));
    	memset(c4, 0, sizeof(c2));
    	int num = 0;
    	//int flag1 = 0, flag2 = 0;
    	for (int i = 0; i < n; i++) {
    		if (a[i].d == 'r' && a[i].c == 'h' && c3[a[i].m] == 0) {
    			b[num++] = i;
    			c3[a[i].m] ++;
    
    		}
    		else if (a[i].d == 'r' && a[i].c == 'a' && c4[a[i].m] == 0) {
    			b[num++] = i;
    			c4[a[i].m] ++;
    
    		}
    			
    		else if (a[i].d == 'y' && c1[a[i].m] == 1 &&  a[i].c == 'h' && c3[a[i].m] == 0) {
    			b[num++] = i;
    			c3[a[i].m] ++;
    		}
    		else if (a[i].d == 'y' && c2[a[i].m] == 1 && a[i].c == 'a' && c4[a[i].m] == 0) {
    			b[num++] = i;
    			c4[a[i].m] ++;
    		}
    		else if (a[i].d == 'y' && a[i].c == 'h')
    			c1[a[i].m] ++;
    		else if (a[i].d == 'y' && a[i].c == 'a')
    			c2[a[i].m] ++;
    	}
    	for (int i = 0; i < num; i++) {
    		if (a[b[i]].c == 'h')
    			cout << str1 << " " << a[b[i]].m << " " << a[b[i]].t << endl;
    		else
    			cout << str2 << " " << a[b[i]].m << " " << a[b[i]].t << endl;
    	}
    	return 0;
    }
    


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  • 原文地址:https://www.cnblogs.com/Tovi/p/6194823.html
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