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Tarjan-求割边的数量
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Tarjan-求割边的数量
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#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e5;
int m = 0,n = 0;
vector<int> g[maxn+10];
bool vis[maxn+10];
int low[maxn+10],dfn[maxn+10];
int cnt = 0,dig = 0;
//这里是tarjan代码模板核心
void tarjan(int u,int w){
vis[u] = true;
dfn[u] = low[u] = (++cnt);
int size = g[u].size();
for(int i = 0;i < size;++i){
int v = g[u][i];
if(v == w){//因为是无向图,防止向后返回去遍历
continue;
}
if(dfn[v] == -1){//没有遍历过,就接着向下遍历,直到找到后,再执行下面的代码,也就是回溯
tarjan(v,u);
low[u] = min(low[u],low[v]);
}else{
low[u] = min(low[u],dfn[v]);
}
}
if(low[u] == dfn[u]){//这个用来求割边的个数,但是无法求割边具体是什么
++dig;//求有几个强连通分支,那么割边的数量等于强连通分支数量-1
}
}
void init(){//初始化函数
memset(vis,false,sizeof(vis));
memset(low,0,sizeof(low));
memset(dfn,-1,sizeof(dfn));
}
int main(){
std::ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
cin >> n >> m;
init();
for(int i = 1;i <= m;++i){
int x = 0,y = 0;
cin >> x >> y;
g[x].push_back(y);g[y].push_back(x);
}
tarjan(1,-1);
cout << dig-1 << endl;//割边的数量
return 0;
}