CF 118E Bertown roads 桥

118E Bertown roads

题目：把无向图指定边的方向，使得原图变成有向图，问能否任意两点之间互达

分析：显然如果没有桥的话，存在满足题意的方案。输出答案时任意从一个点出发遍历一遍即可。

求桥的话，利用tarjan算法的low和dfn值判断一下即可。

#include <set>

#include <map>

#include <list>

#include <cmath>

#include <queue>

#include <stack>

#include <string>

#include <vector>

#include <cstdio>

#include <cstring>

#include <iostream>

#include <algorithm>



using namespace std;



typedef long long ll;

typedef unsigned long long ull;



#define debug puts("here")

#define rep(i,n) for(int i=0;i<n;i++)

#define rep1(i,n) for(int i=1;i<=n;i++)

#define REP(i,a,b) for(int i=a;i<=b;i++)

#define foreach(i,vec) for(unsigned i=0;i<vec.size();i++)

#define pb push_back

#define RD(n) scanf("%d",&n)

#define RD2(x,y) scanf("%d%d",&x,&y)

#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define RD4(x,y,z,w) scanf("%d%d%d%d",&x,&y,&z,&w)

#define All(vec) vec.begin(),vec.end()

#define MP make_pair

#define PII pair<int,int>

#define PQ priority_queue

#define cmax(x,y) x = max(x,y)

#define cmin(x,y) x = min(x,y)

#define Clear(x) memset(x,0,sizeof(x))

/*



#pragma comment(linker, "/STACK:1024000000,1024000000")



int size = 256 << 20; // 256MB

char *p = (char*)malloc(size) + size;

__asm__("movl %0, %%esp\n" :: "r"(p) );



*/



/******** program ********************/



const int MAXN = 1e6+5;



int dfn[MAXN],low[MAXN],dep;

int po[MAXN],tol;

int n,m;



struct node{

    int x,y,id,next;

}edge[MAXN*2];



bool dfs(int x,int fa){

    low[x] = dfn[x] = ++ dep;

    for(int i=po[x];i;i=edge[i].next){

        int y = edge[i].y;

        if(y==fa)continue;

        if(!dfn[y]){

            if(!dfs(y,x))

                return false;

            cmin( low[x],low[y] );



            if(low[y]>dfn[x])

                return false;

        }else

            cmin( low[x],dfn[y] );

    }

    return true;

}



void out(int x,int fa){

    low[x] = 1;

    for(int i=po[x];i;i=edge[i].next){

        int y = edge[i].y;

        if(y==fa)continue;

        //cout<<"dsa "<<x<<" "<<y<<endl;

        if( abs(edge[i].id)==1){

            edge[i].id = 2;

            edge[i^1].id = -2;

        }

        if(!low[y])

            out(y,x);

    }

}



void add(int x,int y,int id){

    edge[++tol].y = y;

    edge[tol].x = x;

    edge[tol].id = id;

    edge[tol].next = po[x];

    po[x] = tol;

}



int main(){



#ifndef ONLINE_JUDGE

    freopen("sum.in","r",stdin);

    //freopen("sum.out","w",stdout);

#endif



    while(cin>>n>>m){

        Clear(po);

        tol = 1;



        int x,y;

        rep1(i,m){

            RD2(x,y);

            add(x,y,1);

            add(y,x,-1);

        }



        Clear(dfn);

        dep = 0;

        bool ok = true;

        rep1(x,n)

            if(!dfn[x]){

                if(!dfs(x,0)){

                    ok = false;

                    break;

                }

            }





        if(ok){

            Clear(low);

            out(1,0);

            for(int i=2;i<=tol;i++)

                if(edge[i].id>0)

                    printf("%d %d\n",edge[i].x,edge[i].y);

        }

        else

            puts("0");

    }



    return 0;

}