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; }