POJ1515Street Direction(无向图改有向图)

题意：
把尽可能多的无向边改成有向边，使得图依然为强连通图

思路：
首先桥必是无向边，不能变。桥左右连接的是强连通分量，对每个强连通分量跑dfs，确定每条边的放方向

#include<iostream>
#include<cstring>
#define N 100001
#define LL long long
using namespace std;
struct Edge{
    int from,to,next;
    int cut;//记录该边是否为桥
    Edge(){}
    Edge(int from,int to,int next,int cut):from(from),to(to),next(next),cut(cut){}
}edge[N*2];//双向边edge[i]与edge[i^2]是两条反向边
int n,m;
int head[N],edgeNum;
int dfn[N],low[N],block_cnt;
bool vis[N];//dfs确定边的方向时的标志数组
bool sig;//判断是否连通
void addEdge(int from,int to){//添边
    edge[edgeNum]=Edge(from,to,head[from],false);
    head[from]=edgeNum++;
}
void tarjin(int x,int father){
	low[x]=dfn[x]=++block_cnt;
	for(int i=head[x];i!=-1;i=edge[i].next)	{
		int y=edge[i].to;
		if(y==father)continue;
		if(!dfn[y]){
			tarjin(y,x);
			if(low[x]>low[y])
                low[x]=low[y];
			if(low[y]>dfn[x])
				edge[i].cut=edge[i^1].cut=true;
		}
		else if(low[x]>dfn[y])
            low[x]=dfn[y];
	}
}
void dfs(int x, int father){
	vis[x]=true;
	for(int i=head[x];~i;i=edge[i].next){
		int y=edge[i].to;
		if(edge[i].cut==1)
            continue;
		if(edge[i^1].cut!=-1)
           edge[i].cut=-1;
		if(!vis[y])
			dfs(y,x);
	}
}
int main()
{
    int r=0;
    while(scanf("%d%d",&n,&m)!=EOF&&(n+m)){
        memset(dfn,0,sizeof(dfn));
        memset(head,-1,sizeof(head));
        memset(vis,0,sizeof(vis));
        edgeNum=0;
        while(m--){
            int x,y;
            scanf("%d%d",&x,&y);
            addEdge(x,y);
            addEdge(y,x);
        }
        tarjin(1,1);
        for(int i=1;i<=n;i++)
            if(!vis[i])
                dfs(i,i);
        printf("%d\n\n",++r);
        for(int i=0;i<edgeNum;i++)
            if(edge[i].cut)
                printf("%d %d\n",edge[i].from,edge[i].to);
        printf("#\n");
    }
    return 0;
}