TOJ 2933 ZOJ 1589 Professor John / floyd

Professor John

时间限制(普通/Java):1000MS/3000MS 运行内存限制:65536KByte



描述

Professor John is investigating a math problem. He has obtained some relations among several variables. Now he would like to know if there are any other relations that can be deduced
from these obtained ones. Since he has been working for too long, Professor John decides to grant himself a vacation while assigning you to do the job. Are you ready?

输入

The first line of input contains an integer N, which is the number of test cases. Then N test cases follow.

For each test case:

the 1st line contains a positive integer m (<= 100) which is the number of given relations;

the following m lines each contains a given relation, in the format

Variable1<Variable2

or

Variable1>Variable2

A "Variable" is represented by a capital character. There will not be conflicting relations given.

输出

For each test case, first print in one line "Case d:" where d is the number of the test case, start counting from 1.

Then output all the relations which can be deduced from the given relations in alphabetical order, in the format Variable1<Variable2. Each relation occupies one line. No extra space shall be printed. The given relations must NOT be included.

If no new relation is found, output "NONE" in one line.

样例输入

[code]2
3
A<strong c="">B
C<D
2
A<B
C<D
</strong>

样例输出

[code]Case 1:
A<C
A<D
B<D
Case 2:
NONE

floyd求传递闭包

#include<stdio.h>
#include<string.h>
#define inf 0x7fffffff
const int MAX = 33;
int a[MAX][MAX];
int flag[MAX][MAX];
void floyd()
{
	int i,j,k;
	for(k = 1; k <= 30; k++)
		for(i = 1; i <= 30; i++)
			for(j = 1; j <= 30; j++)
				if(a[i][k] && a[k][j])
					a[i][j] = 1;
}
int main()
{
	char ss[5];
	int t,i,m,j,f,cas = 1;
	scanf("%d",&t);
	while(t--)
	{
		f = 0;
		memset(a,0,sizeof(a));
		memset(flag,0,sizeof(flag));
		scanf("%d",&m);
		while(m--)
		{
			scanf("%s",ss);
			if(ss[1] == '<')
			{
				a[ss[0]-'A'+1][ss[2]-'A'+1] = 1;
				flag[ss[0]-'A'+1][ss[2]-'A'+1] = 1;
			}
			else
			{
				a[ss[2]-'A'+1][ss[0]-'A'+1] = 1;
				flag[ss[2]-'A'+1][ss[0]-'A'+1] = 1;
			}
		}
		floyd();
		printf("Case %d:\n",cas++);
		for(i = 1;i <= 30; i++)
			for(j = 1;j <= 30; j++)
				if(a[i][j] && !flag[i][j])
				{
					printf("%c<%c\n",i+'A'-1,j+'A'-1);
					f++;
				}
		if(!f)
			printf("NONE\n");
	}
	return 0;
}