poj 2570 Fiber Network(传递闭包，floyd+位运算)

[align=center]FiberNetwork[/align]

TimeLimit:1000MS		MemoryLimit:65536K
TotalSubmissions:2703		Accepted:1242

Description
SeveralstartupcompanieshavedecidedtobuildabetterInternet,calledthe"FiberNet".Theyhavealreadyinstalledmanynodesthatactasroutersallaroundtheworld.Unfortunately,theystartedtoquarrelabouttheconnecting
lines,andendedupwitheverycompanylayingitsownsetofcablesbetweensomeofthenodes.

Now,serviceproviders,whowanttosenddatafromnodeAtonodeBarecurious,whichcompanyisabletoprovidethenecessaryconnections.Helptheprovidersbyansweringtheirqueries.
Input
Theinputcontainsseveraltestcases.Eachtestcasestartswiththenumberofnodesofthenetworkn.Inputisterminatedbyn=0.Otherwise,1<=n<=200.Nodeshavethenumbers1,...,n.Thenfollowsalistofconnections.Every
connectionstartswithtwonumbersA,B.ThelistofconnectionsisterminatedbyA=B=0.Otherwise,1<=A,B<=n,andtheydenotethestartandtheendpointoftheunidirectionalconnection,respectively.Foreveryconnection,thetwonodesarefollowedbythe
companiesthathaveaconnectionfromnodeAtonodeB.Acompanyisidentifiedbyalower-caseletter.Thesetofcompanieshavingaconnectionisjustawordcomposedoflower-caseletters.

Afterthelistofconnections,eachtestcaseiscompletedbyalistofqueries.EachqueryconsistsoftwonumbersA,B.Thelist(andwithitthetestcase)isterminatedbyA=B=0.Otherwise,1<=A,B<=n,andtheydenotethestartandtheendpointofthequery.
Youmayassumethatnoconnectionandnoquerycontainsidenticalstartandendnodes.
Output
Foreachqueryineverytestcasegeneratealinecontainingtheidentifiersofallthecompanies,thatcanroutedatapackagesontheirownconnectionsfromthestartnodetotheendnodeofthequery.Iftherearenocompanies,
output"-"instead.Outputablanklineaftereachtestcase.


SampleInput

3
12abc
23ad
13b
31de
00
13
21
32
00
2
12z
00
12
21
00
0

SampleOutput

ab
d
-

z
-

题意：给出n个结点，给出若干条连接结点的路径，路径是单向的，路径上有小写字母。例如：当结点1和结点2之间的路径有a、b，结点2和结点3之间的路径有b、c，那么结点1和结点3之间的路径就会出现b（即使结点1和结点3之间没有直接相连的路径或路径上没有b）。每次询问结点a和b之间的路径有哪个小写字母。

思路：传递闭包，用floyd。若n^3的floyd再加上查找相同字母，则显然TLE。有什么方法不用查找相同的字母？用位运算！G[a][b]保存a与b路径上的小写字母的集合，通过并（|）运算即可合并结合

AC代码：

#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<cmath>
#include<vector>
#include<cstdlib>
#include<iostream>

usingnamespacestd;

intmain()
{
intn;
inta,b;
intG[205][205];
charcom[30];
while(cin>>n,n)
{
memset(G,0,sizeof(G));
while(cin>>a>>b)
{
if(!a&&!b)break;
cin>>com;
inti=0;
while(com[i])
G[a][b]|=1<<(com[i++]-'a');
}
for(intk=1;k<=n;k++)
for(inti=1;i<=n;i++)
for(intj=1;j<=n;j++)
if(G[i][k]&G[k][j])
G[i][j]|=G[i][k]&G[k][j];

while(cin>>a>>b)
{
if(!a&&!b)break;
if(G[a][b]!=0)
{
inti=0,temp=G[a][b];
while(temp)
{
if(temp&1)
putchar(i+'a');
temp>>=1;
i++;
}
}
else
putchar('-');
putchar('\n');
}
putchar('\n');
}
return0;
}