Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 125536/65536 K (Java/Others)
Total Submission(s): 2885 Accepted Submission(s):
726
#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;
}
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】
原文:http://www.cnblogs.com/tonghao/p/4869529.html