zoj2588 (non-direction graph for cutting edge)

This problem is actually a very simple cutting edge template problem, but need to deal with the re-edge of the tutor's opinion a bit troublesome things

#include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> using namespace
Std
const int maxn=20005;
const int maxm=200005;
struct Edge {int to,next,id,num;//This place requires an ID (the label of the Edge) and num (whether heavy-edged) bool cut on the base of the original template;} EDGE[MAXM];
int ANS[MAXM];
int Head[maxn],tot;
int LOW[MAXN],DFN[MAXN],STACK[MAXN];
int index,top;
BOOL INSTACK[MAXN];
BOOL CUT[MAXN];
int ADD_BLOCK[MAXN];
int bridge;
int t,a,b;
Long Long n,m;
    void Addedge (int u,int v,int x,int num)//I was right when I passed four values, did not process {edge[tot].to=v When I did not pass Num;
    Edge[tot].next=head[u];
    Edge[tot].cut=false;
    Edge[tot].id=x;
    Edge[tot].num=num;
head[u]=tot++;
    } void Tarjan (int u,int pre) {int v,k=1;
    Low[u]=dfn[u]=++index;
    Stack[top++]=u;
    Instack[u]=true;
    int son=0,sum=0;
        for (int i=head[u]; i!=-1; i=edge[i].next) {v=edge[i].to;
        if (v==pre) continue;
            if (!dfn[v]) {son++;
            Tarjan (V,u); If(Low[u]>low[v]) low[u]=low[v];
                if (low[v]>dfn[u]&&edge[i].num!=1) {bridge++;
                Edge[i].cut=true;
                Edge[i^1].cut=true;
    ans[bridge]=edge[i].id;//here is directly the label of the Edge}} else if (Low[u]>dfn[v]) low[u]=dfn[v];
    } if (U==pre) add_block[u]=son-1;
    Instack[u]=false;
top--;
    } void Init () {tot=0;
    bridge=0;
    index=0;
    memset (low,0,sizeof (Low));
    memset (dfn,0,sizeof (DFN));
    memset (instack,0,sizeof (instack));
    memset (add_block,0,sizeof (Add_block));
    memset (Edge,-1,sizeof (Edge));
    memset (head,-1,sizeof (head));
memset (ans,0,sizeof (ans));
    } int main () {Ios::sync_with_stdio (false);
    cin>>t;
        while (t--) {init ();
        cin>>n>>m;
            for (int i=1; i<=m; i++) {cin>>a>>b;
for (int j=head[a]; j!=-1; j=edge[j].next)                if (edge[j].to==b) {edge[i].num=1;
            } addedge (A,b,i,edge[i].num);
        Addedge (B,a,i,edge[i].num);
        } for (int i=1; i<=n; i++) {Tarjan (i,i); } cout<<bridge<<endl;//the output to be processed, the first need to consider the problem of output empty line, and then need to consider the absence of a bridge, here eat a good loss, has been WA to dead sort (ans+1,an
        S+BRIDGE+1);
              for (int i=1,u=bridge; i<=bridge; i++) {cout<<ans[i];
         if (--u) cout<< "";
            } if (bridge) cout<<endl;
    if (T) cout<<endl;
} return 0;
 }