UVA - 10420 - List of Conquests （排序 - qsort、STL）

UVA - 10420

List of Conquests

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

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Description

Problem B
List of Conquests
Input: standard input
Output: standard output
Time Limit: 2 seconds

In Act I, Leporello is telling Donna Elvira about his master's long list of conquests:

``This is the list of the beauties my master has loved, a list I've made out myself: take a look, read it with me. In Italy six hundred and forty, in Germany two hundred and thirty-one, a hundred in France, ninety-one in Turkey; but in Spain already a thousand and three! Among them are country girls, waiting-maids, city beauties; there are countesses, baronesses, marchionesses, princesses: women of every rank, of every size, of every age.'' (Madamina, il catalogo è questo)

As Leporello records all the ``beauties'' Don Giovanni ``loved'' in chronological order, it is very troublesome for him to present his master's conquest to others because he needs to count the number of ``beauties'' by their nationality each time. You are to help Leporello to count.

Input

The input consists of at most 2000 lines, but the first. The first line contains a number n, indicating that there will be n more lines. Each following line, with at most 75 characters, contains a country (the first word) and the name of a woman (the rest of the words in the line) Giovanni loved. You may assume that the name of all countries consist of only one word.

Output

The output consists of lines in alphabetical order. Each line starts with the name of a country, followed by the total number of women Giovanni loved in that country, separated by a space.

Sample Input

3 

Spain Donna Elvira 

England Jane Doe 

Spain Donna Anna 

Sample Output

England 1 

Spain 2

Problem-setter: Thomas Tang, Queens University, Canada

 

 

“Failure to produce a reasonably good and error free problem set illustrates two things a) Lack of creativity b) Lack of commitment”

Source

Root :: AOAPC I: Beginning Algorithm Contests (Rujia Liu) :: Volume 1. Elementary Problem Solving ::  Sorting/Searching
Root :: Competitive Programming: Increasing the Lower Bound of Programming Contests (Steven & Felix Halim) :: Chapter 1. Introduction ::  Ad Hoc
Root :: Competitive Programming 3: The New Lower Bound of Programming Contests (Steven & Felix Halim) :: String Processing :: Ad Hoc String Processing Problems ::  Frequency Counting
Root :: Competitive Programming 2: This increases the lower bound of Programming Contests. Again (Steven & Felix Halim) :: Introduction :: Ad Hoc Problems - Part 1 ::  Easy

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思路：一个人在多个国家有爱人，输入每一组包括一个国家和爱人，算出每个国家的爱人数并输出，输出按照字典序

AC代码①（qsort的做法）：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath> 
using namespace std;

char na[100], name[100];

struct node {
	char name[100];
	int cnt;
};

struct node nation[2005];

int cmp(const void *a, const void *b) {
	return strcmp(((node*)a)->name, ((node*)b)->name);
}

int main() {
	int T, num = 0;
	scanf("%d", &T);
	while(T--) {
		scanf("%s", na);
		gets(name);
		
		int flag = 0;
		for(int i = 0; i < num; i ++) {
			if(strcmp(na, nation[i].name) == 0) {
				flag = 1;
				nation[i].cnt++;
				break;
			}
		}
		if(!flag) {
			nation[num].cnt = 1;
			strcpy(nation[num].name, na);
			num++;
		}
	}
	qsort(nation, num, sizeof(node), cmp);
	for(int i = 0; i < num; i++) {
		printf("%s %d\n", nation[i].name, nation[i].cnt);
	}
	return 0;
}

AC代码②（STL中map的做法）：

#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <algorithm>
#include <cmath> 
#include <map>
using namespace std;

char na[100], tmp[100];

map<string, int> nation;

int main() {
	int T, num = 0;
	scanf("%d", &T);
	while(T--) {
		scanf("%s", na);
		gets(tmp);
		if(nation.count(na)) nation[na]++;
		else nation[na] = 1;
	}
	for(map<string, int> :: iterator it = nation.begin(); it != nation.end(); it++) {
		cout << it->first << ' ' << it->second <<endl;
	}
	return 0;
}

AC代码③（STL中map的做法2，注意输出）：

#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath> 
#include <map>
using namespace std;

char na[100], tmp[100];

map<string, int> nation;

int main() {
	int T, num = 0;
	scanf("%d", &T);
	while(T--) {
		scanf("%s", na);
		gets(tmp);
		if(nation.count(na)) nation[na]++;
		else nation[na] = 1;
	}
	for(map<string, int> :: iterator it = nation.begin(); it != nation.end(); it++) {
		printf("%s %d\n", it->first.c_str(), it->second);
	}
	return 0;
}

AC代码④（sort，待续。。）：