Figure Unicom problem

Cutting edge (Bridge)

template

vector<pair<int, int> > ans;
vector<int> e[maxn];
void dfs(int u,int fa) {
    dep[u] = dep[fa] + 1;
    for(int v:e[u]) {
        if(v==fa)
            continue;
        if(!dep[v]) {
            dfs(v, u);
            dp[u] += dp[v];
        }
        else {
            if(dep[v]>dep[u]) {
                dp[u]--;//down from u
            }
            else if(dep[v]<dep[u]) {
                dp[u]++;//up form u
            }
        }
    }
    if(!dp[u]&&u!=1){
        brige = 1;
    }
} 

\(dp[u]\)= (#of back-edges going up from u) ​− (# of back-edges going down from u) +\(\sum dp[v]\)

If a point does not back-edge, then indicating the presence of the bridge of FIG.

example

Bertown roads

CF118E

The meaning of problems

FIG n points to an m edges, undirected edges; To this now to FIG become edges having unidirectional FIG. M, and guaranteed to arrive one another between any two points

Thinking

DFS tree, and count the number of back-edge of each point, if it is 0 it represents the parent form cutting bridge, to the output 0

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