[USACO06JAN]冗余路径Redundant Paths 无向图tarjan缩点

如题，缩完点后数一下有几个入度为1的scc，+1再/2即可。

教训：加一个cntf处理重边！否则重边会被认为是同一条。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;

struct stack{
    vector<int> v;
    void push(int x){v.push_back(x);}
    void pop(){v.pop_back();}
    int size(){return v.size();}
    bool empty(){return v.empty();}
    int top(){return v[v.size()-1];}
}s;

const int Maxn = 5010;

vector<int> g[Maxn],scc[Maxn];
int dfn[Maxn],low[Maxn],inscc[Maxn],ins[Maxn];
int inv[Maxn];
int n,m,cntv,cntscc,cnt;

void tarjan(int x,int fa){
    low[x] = dfn[x] = ++cntv,ins[x] = 1;
    s.push(x);int cntf = 0;
    for(int i = 0;i < g[x].size();i++){
        int u = g[x][i];
        if(u == fa&&!cntf){cntf++;continue;}
        if(!dfn[u])tarjan(u,x),low[x] = min(low[x],low[u]);
        else low[x] = min(low[x],dfn[u]);
    }
    if(dfn[x] == low[x]){
        int y = -1;cntscc++;
        while(!s.empty()&&y^x){
            y = s.top();s.pop();
            ins[y] = 0,inscc[y] = cntscc;
            scc[cntscc].push_back(y);
        }
    }
}

int main(){
    scanf("%d%d",&n,&m);
    for(int i = 1;i <= m;i++){
        int u,v;
        scanf("%d%d",&u,&v);
        g[u].push_back(v);
        g[v].push_back(u);
    }
    tarjan(1,-1);
    for(int i = 1;i <= n;i++)
        for(int j = 0;j < g[i].size();j++)
            if(inscc[i]^inscc[g[i][j]])inv[inscc[g[i][j]]]++;
    for(int i = 1;i <= cntscc;i++)if(inv[i] == 1)cnt++;
    cout << (cnt+1)/2;
return 0;
}