POJ 3352 Road Construction (conclusion of adding a double connected graph with the least side construction edge)

Test instructions: Known undirected graph, ask to add the least edge to make it a side double connected graph

Idea: It is obvious to indent the point into a tree, add the least side to make the edge of a tree connected

Here is the conclusion: for a tree to add (1+leaf) >>1 strip without the edge can be structured into a double-connected graph, the construction method is obvious (brain repair

    216K 63MS C + + 1754B #include <cstdio> #include <iostream> #include <cstring> #include

    <algorithm> using namespace std;
    const int N = 1010;
    struct node {int v,next;
    }es[n*n];
    int head[n];
    int n,m;
    int low[n],dfn[n],index;

    int indeg[n];
        void Ini () {memset (head,-1,sizeof (head));
        memset (dfn,0,sizeof (DFN));
        index=0;
    memset (indeg,0,sizeof (indeg));
        } void Tarjan (int u,int pa) {dfn[u]=low[u]=++index;
            for (int i=head[u];~i;i=es[i].next) {int v=es[i].v;
                if (dfn[v]==0) {Tarjan (v,u);
            Low[u]=min (Low[u],low[v]);
        } if (V!=PA) low[u]=min (Low[u],low[v]);
            }} int main () {while (~scanf ("%d%d", &n,&m)) {ini ();
         for (int i=1;i<=m;i++) {int u,v;       scanf ("%d%d", &u,&v);
                Es[i].v=v;
                Es[i].next=head[u];
                Head[u]=i;
                Es[i+m].v=u;
                ES[I+M].NEXT=HEAD[V];
            Head[v]=i+m;

            } for (int i=1;i<=n;i++) if (dfn[i]==0) Tarjan (i,-1);
                    for (int u=1;u<=n;u++) {for (int i=head[u];~i;i=es[i].next) {
                    int v= es[i].v;
                    if (Low[v]==low[u]) continue;
                indeg[low[u]]++;
            }} int leaf=0;
            for (int i=1;i<=n;i++) if (indeg[i]==1) leaf++;

        printf ("%d\n", (leaf+1)/2);
    } return 0;
 }

The above code is a self-adaptation of a Tarjan, only in this problem can be used, has not proved a clear error, the following is the positive solution:

#include <cstdio> #include <cstring> #include <iostream> #include <cstring> using namespace std;
int n,m;
const int N = 1000+100;
const int M = 2*n;
int head[n],cnt; struct Edge {int v,nxt;}

ES[M];
    inline void Add_edge (int u,int v) {es[cnt].v=v;
    Es[cnt].nxt=head[u];
    head[u]=cnt++;
    Es[cnt].v=u;
    ES[CNT].NXT=HEAD[V];
head[v]=cnt++;
} typedef pair<int,int>pii;
PII Bridge[m];
int dfn[n],sta[n],top,scc,index,col[n],low[n],cbr;
    void Tarjan (int u,int pa) {dfn[u]=low[u]=++index;
    Sta[++top]=u;
        for (int i=head[u];~i;i=es[i].nxt) {int v=es[i].v;
        if (V==PA) continue;
            if (dfn[v]==0) {Tarjan (v,u);
            Low[u]=min (Low[u],low[v]);
                if (Low[v]>dfn[u]) {scc++;
                int VV;
                    do {vv=sta[top--];
                COL[VV]=SCC;
                }while (VV!=V); bridge[++cbr]=pii (U,V);
    }} else Low[u]=min (Low[u],dfn[v]);
}} int in[n];
    void Ini () {memset (dfn,0,sizeof (DFN));
    memset (col,0,sizeof (col));
    memset (In,0,sizeof (in));
    memset (head,-1,sizeof (head));
top=scc=cbr=index=cnt=0;
        } int main () {while (~scanf ("%d%d", &n,&m)) {ini ();
            for (int i=1;i<=m;i++) {int u,v;
            scanf ("%d%d", &u,&v);
        Add_edge (U,V);
        } for (int i=1;i<=n;i++) if (!dfn[i]) Tarjan (i,-1);
        if (top) scc++;
        while (top) {COL[STA[TOP--]]=SCC;
            } for (int i=1;i<=cbr;i++) {int u=bridge[i].first,v=bridge[i].second;
            in[col[u]]++;
        in[col[v]]++;
        } int leaf=0;
        for (int i=1;i<=scc;i++) if (in[i]==1) leaf++;
    printf ("%d\n", (leaf+1) >>1);

} return 0;

 }