uva 10044 Erdös Numbers

Erdös Numbers 解题报告

最新推荐文章于 2019-01-16 13:53:33 发布

原创最新推荐文章于 2019-01-16 13:53:33 发布 · 840 阅读

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#numbers #struct #list #测试

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该博客介绍了如何解决UVA 10044题目——Erdös Numbers。通过建立数学家之间的合作网络，计算每个数学家的Erdös Number，即与Erdös合作的论文间隔数。博客中提供了C语言的解决方案，包括数据结构和算法实现，对大型输入数据进行了处理。

题意：有一个数学家名字是Erdos P. 与他一起发表过论文的人的Erdös Number 是 1 与Erdös Number是1的人发表过论文的是2 依次类推木有Erdös Number数的输出

infinity

注意测试数据很强也很大

连接点击打开链接

#include <stdio.h>
#include <string.h>
#define MAX 0x3fffffff
#define Max 5000
struct list
{
	char name[1000];
	int num;
}s[Max];

bool visit[Max][Max];

void initialize ( )
{
	int i, j;
	for ( i = 0; i < Max; i++ )
		for ( j = 0; j < Max; j++ )
			visit[i][j] = false;
	strcpy ( s[0].name, "Erdos, P." );
	s[0].num = 0;
	for ( i = 1; i < Max; i++ )
		s[i].num = MAX;
}

void solve ( int n, int m )
{
	int sum = 1, i, j, k;
	char str[100000];
	initialize ( );
	for ( i = 0; i < n; i++ )
	{
		gets ( str );
		char ch[1000];
		bool flag[Max] = { false };
		int x = 0, len = strlen ( str );
		for ( j = 0; j < len; j++ )
		{
			if ( x != 0 || str[j] != ' ' )
				ch[x++] = str[j];
			if ( ( j - 1 > 0 && str[j] == ',' && str[j-1] == '.' ) || str[j] == ':' )
			{
				x--;
				ch[x] = '\0';
				bool find = false;
				for ( k = 0; k < sum; k++ )
					if ( !strcmp ( ch, s[k].name ) )
					{
						find = true;
						x = k;
						break;
					}
				if ( !find )
				{
					strcpy ( s[sum].name, ch );
					x = sum;
					sum++;
				}
				flag[x] = true;
				x = 0;
				if ( str[j] == ':' )
					break;
			}
		}
		for ( j = 0; j < sum; j++ )
			if ( flag[j] )
				for ( k = 0; k < sum; k++ )
					if ( flag[k] && k != j )
						visit[j][k] = true;
	}
	for ( i = 0; i < sum; i++ )
		for ( j = 0; j < sum; j++ )
			if ( s[j].num == i )
			{
				for ( k = 0; k < sum; k++ )
					if ( visit[j][k] )
					{
						if ( s[k].num > i + 1 )
							s[k].num = i + 1;
					}
			}
	for ( i = 0; i < m; i++ )
	{
		gets ( str );
		for ( j = 0; j < sum; j++ )
			if ( !strcmp ( str, s[j].name ) )
				break;
		printf ( "%s ", str );
		if ( s[j].num == MAX )
			puts ( "infinity" );
		else
			printf ( "%d\n", s[j].num );
	}
}

int main ( )
{
	int n, m, t;
	scanf ( "%d", &t );
	for ( int i = 1; i <= t; i++ )
	{
		printf ( "Scenario %d\n", i );
		scanf ( "%d%d\n", &n, &m );
		solve ( n, m );
	}
	return 0;
}