hdoj 3849 By Recognizing These Guys, We Find Social Networks Useful【双连通分量求桥&&输出桥&&字符串处理】

By Recognizing These Guys, We Find Social Networks Useful

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 125536/65536 K (Java/Others)
Total Submission(s): 2885    Accepted Submission(s):
726

Problem Description

Social Network is popular these days.The Network helps
us know about those guys who we are following intensely and makes us keep up our
pace with the trend of modern times.
But how?
By what method can we know
the infomation we wanna?In some websites,maybe Renren,based on social network,we
mostly get the infomation by some relations with those "popular leaders".It
seems that they know every lately news and are always online.They are alway
publishing breaking news and by our relations with them we are informed of
"almost everything".
(Aha,"almost everything",what an impulsive
society!)
Now,it's time to know what our problem is.We want to know which are
the key relations make us related with other ones in the social
network.
Well,what is the so-called key relation?
It means if the relation
is cancelled or does not exist anymore,we will permanently lose the relations
with some guys in the social network.Apparently,we don't wanna lose relations
with those guys.We must know which are these key relations so that we can
maintain these relations better.
We will give you a relation description map
and you should find the key relations in it.
We all know that the relation
bewteen two guys is mutual,because this relation description map doesn't
describe the relations in twitter or google+.For example,in the situation of
this problem,if I know you,you know me,too.

 

Input

The input is a relation description map.
In the
first line,an integer t,represents the number of cases(t <= 5).
In the
second line,an integer n,represents the number of guys(1 <= n <= 10000)
and an integer m,represents the number of relations between those guys(0 <= m
<= 100000).
From the second to the (m + 1)the line,in each line,there are
two strings A and B(1 <= length[a],length[b] <= 15,assuming that only
lowercase letters exist).
We guanrantee that in the relation description
map,no one has relations with himself(herself),and there won't be identical
relations(namely,if "aaa bbb" has already exists in one line,in the following
lines,there won't be any more "aaa bbb" or "bbb aaa").
We won't guarantee
that all these guys have relations with each other(no matter directly or
indirectly),so of course,maybe there are no key relations in the relation
description map.

 

Output

In the first line,output an integer n,represents the
number of key relations in the relation description map.
From the second line
to the (n + 1)th line,output these key relations according to the order and
format of the input.

 

Sample Input

1

4 4

saerdna aswmtjdsj

aswmtjdsj mabodx

mabodx biribiri

aswmtjdsj biribiri

 

Sample Output

1

saerdna aswmtjdsj

 

第一次直接用数组处理字符串  超时了   这是超时代码

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<stack>
#define MAX 100100
#define INF 0x7fffff
using namespace std;
int n,m;
int head[MAX],ans;
int dfn[MAX],low[MAX];
int dfsclock;
char str1[50],str2[50];
char p[MAX][20];
int num,mark;
char ant[MAX][20],ano[MAX][20];
struct node
{
	int beg,end,next;
}edge[MAX];
void init()
{
	ans=num=0;
	memset(head,-1,sizeof(head));
}
void add(int u,int v)
{
	node E={u,v,head[u]};
	edge[ans]=E;
	head[u]=ans++;
}
void getmap()
{
	int i,j,a,b;
	int sum=1;
	for(i=1;i<=m;i++)
	{
		scanf("%s%s",str1,str2);
		int x,y;
		x=y=INF;
		a=b=0;
		for(j=1;j<sum;j++)
		{
			if(strcmp(str1,p[j])==0)
		    {
		    	x=j;
		    	a=j;
		    }
			if(strcmp(str2,p[j])==0)
			{
				y=j;
				b=j;
			}
			if(x!=INF&&y!=INF)
			    break;
		}
		if(x==INF)
		{
			x=sum++;
			a=x;
			strcpy(p[x],str1);
		}
		if(y==INF)
		{
			y=sum++;
			b=y;
			strcpy(p[y],str2);
		}
		add(a,b);
		add(b,a);
	}
}
void tarjan(int u,int fa)
{
	int i,v;
	low[u]=dfn[u]=++dfsclock;
	for(i=head[u];i!=-1;i=edge[i].next)
	{
		v=edge[i].end;
		if(v==fa)
		    continue;
		if(!dfn[v])
		{
			tarjan(v,u);
			low[u]=min(low[u],low[v]);
			if(dfn[u]<low[v])
			{
				strcpy(ant[num],p[edge[i].beg]);
				strcpy(ano[num++],p[edge[i].end]);
			}
		}
		else
		    low[u]=min(low[u],dfn[v]);
	}
}
void find()
{
	memset(low,0,sizeof(low));
	memset(dfn,0,sizeof(dfn));
	dfsclock=0;
	tarjan(1,-1);
	mark=0;
	for(i=1;i<=n;i++)
	{
		if(!dfn[i])
		{
			mark=1;
			break;
		}
	}
}
void solve()
{
	int i,j;
	if(mark)
	{
		printf("0\n");
		return ;
	}
	printf("%d\n",num);
	for(i=0;i<num;i++)
	{
		printf("%s %s\n",ant[i],ano[i]);
	}
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		init();
		getmap();
		find();
		solve();
	}
	return 0;
}

　　后来看别人都是用的map处理  我不会用map 苦逼啊

#include<stdio.h>
#include<string.h>
#include<string>
#include<algorithm>
#include<stack>
#include<map>
#define MAX 200010
#define INF 0x7fffff
using namespace std;
int n,m;
int head[MAX],ans;
int dfn[MAX],low[MAX];
int dfsclock;
char str1[50],str2[50];
map<string, int>q1;
map<int, string>q2;
int num,mark,bridge;
struct node
{
	int beg,end,cnt,next;
}edge[MAX];
void init()
{
	ans=num=bridge=0;
	memset(head,-1,sizeof(head));
}
void add(int u,int v)
{
	node E={u,v,0,head[u]};
	edge[ans]=E;
	head[u]=ans++;
}
void getmap()
{
	int i,j,a,b;
	int sum=1;
	q1.clear();
	q2.clear();
	for(i=1;i<=m;i++)
	{
		scanf("%s%s",str1,str2);
		if(q1[str1]==0)//貌似是因为map中都是一对一的情况，即同一个字符串不可能重复出现
		{
			q1[str1]=sum;
			q2[sum]=str1;
			sum++;
		}
		if(q1[str2]==0)
		{
			q1[str2]=sum;
			q2[sum]=str2;
			sum++;
		}
		add(q1[str1],q1[str2]);
		add(q1[str2],q1[str1]);
	}
}
void tarjan(int u,int fa)
{
	int i,v;
	low[u]=dfn[u]=++dfsclock;
	for(i=head[u];i!=-1;i=edge[i].next)
	{
		v=edge[i].end;
		if(v==fa)
		    continue;
		if(!dfn[v])
		{
			tarjan(v,u);
			low[u]=min(low[u],low[v]);
			if(dfn[u]<low[v])
			{
				edge[i].cnt=edge[i^1].cnt=1;
				bridge++;
			}
		}
		else
		    low[u]=min(low[u],dfn[v]);
	}
}
void find()
{
	memset(low,0,sizeof(low));
	memset(dfn,0,sizeof(dfn));
	dfsclock=0;
	tarjan(1,-1);
	mark=1;
	for(int i=1;i<=n;i++)
	{
		if(!dfn[i])
		{
			mark=0;
			return ;
		}
	}
}
void solve()
{
	int i,j;
	if(mark==0)
	    printf("0\n");
	else
	{
		printf("%d\n",bridge);
		for(i=0;i<ans;i=i+2)
		{
			if(edge[i].cnt==1)
			    printf("%s %s\n",q2[edge[i].beg].c_str(),q2[edge[i].end].c_str());
		}       //我查了map的常用函数，但是没找到这个.c_str  .......
	}
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		init();
		getmap();
		find();
		solve();
	}
	return 0;
}