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.
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.
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.
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
");
return ;
}
printf("%d
",num);
for(i=0;i<num;i++)
{
printf("%s %s
",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
");
else
{
printf("%d
",bridge);
for(i=0;i<ans;i=i+2)
{
if(edge[i].cnt==1)
printf("%s %s
",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;
}