UVA796 Critical Links (tanjar求桥)

来源：互联网发布：矩阵的秩8个性质及证明编辑：程序博客网时间：2024/06/01 19:09

题目大意：给出一张无向图，按顺序输出桥

注意桥输出的时候需要排序

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>using namespace std;#define LL long long#define mem(a,b) memset(a,b,sizeof a)const int INF = 0x3f3f3f3f;#define MAXN 100100#define MAXM 1000100struct node{    int u,v,flag,next;} edge[MAXM];struct node2{    int u,v;} ans[MAXM];int dfn[MAXN],low[MAXN],s[MAXN],fl[MAXM];int cnt,tot,ant;bool cmp(node2 a,node2 b){    if(a.u!=b.u)        return a.u<b.u;    return a.v<b.v;}void init(){    mem(s,-1);    mem(fl,0);    mem(dfn,0);    mem(low,0);    cnt=tot=ant=0;}void add(int u,int v){    for(int i=s[u]; ~i; i=edge[i].next)    {        if(edge[i].v==v)        {            edge[i].flag++;            return;        }    }    edge[cnt].u=u;    edge[cnt].v=v;    edge[cnt].flag=0;    edge[cnt].next=s[u];    s[u]=cnt++;}void tarjan(int x,int f){    dfn[x]=low[x]=++tot;    for(int i=s[x]; ~i; i=edge[i].next)    {        int v=edge[i].v;        if(v==f)            continue;        if(!dfn[v])        {            tarjan(v,x);            low[x]=min(low[x],low[v]);        }        else        {            low[x]=min(low[x],dfn[v]);        }        if(low[v]>dfn[x]&&edge[i].flag==0)        {            ans[ant].u=x;            ans[ant].v=v;            if(x>v)                swap(ans[ant].u,ans[ant].v);            ant++;        }    }}int main(){    int n,m,u,v;    char c;    while(~scanf("%d",&n))    {        init();        for(int i=0; i<n; i++)        {            scanf("%d (%d)",&u,&m);            for(int i=0; i<m; i++)            {                scanf("%d",&v);                add(u,v);            }        }        for(int i=0; i<n; i++)            if(!dfn[i]) tarjan(i,-1);        sort(ans,ans+ant,cmp);        printf("%d critical links\n",ant);        for(int i=0; i<ant; i++)            printf("%d - %d\n",ans[i].u,ans[i].v);        printf("\n");    }    return 0;}

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