你就dfs的时候记录一下出现了多少条连向父亲的边就可以了 然后和有向图不一样的是,在这里非树边的更新不用判断点是不是在栈内,因为无向图中没有u可以到达v但是v不能到达u的情况 //Author: dream_maker#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<stack>using namespaces...
46template<classT> inline T Min(T a,T b){returna<b?a:b;}47template<classT> inline T Max(T a,T b){returna>b?a:b;}48template<classT> inline T Min(T a,T b,T c){returnmin(min(a, b),c);}49template<classT> inline T Max(T a,T b,T c){returnmax(max(a, b),c);}...
b:array[1..maxn,1..maxn] of boolean; indegree,c,a,low:array[1..maxn] of longint; time:longint; function min(a,b:longint):longint; begin if ab then exit(a) else exit(b); end; procedure tarjan(k,father:longint); var i,j:longint; begin inc(time); a[k]:=time; low[...
代码: #include<iostream>#include<cstring>#include<cstdio>usingnamespacestd;#defineMAXN5005#defineMAXM22000structEdge{intto,next;}edge[MAXM];intfirst[MAXN],DFN[MAXN],Low[MAXN];boolmap[MAXN][MAXN];intdegree[MAXN];intcnt,tot,n,m,count;voidTarjan(intv,intpre){DFN[v]=Low[v]=++count...
1/*这是一个63分的代码,因为没有注意到题目中的重边问题,以后要注意有重边的图和没有重边的图的tarjan求桥的算法,是不同的*/2#include<iostream>3usingnamespacestd;4#include<cstdio>5#defineN 50016#defineR 100107#include<stack>8#include<queue>9#include<cstring>10queue<int>que;11boolqiao[R]={...