P2860 [USACO06JAN]冗余路径Redundant Paths

实际上是一道tarjan缩点的题，tarjan求点双联通分量，主要思路就是缩完点之后，找新图中的叶子节点的个数，然后带入公式就行了。详情见lba的博客。

题干：

。。。略。

代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<ctime>
#include<queue>
#include<algorithm>
#include<cstring>
using namespace std;
#define duke(i,a,n) for(int i = a;i <= n;i++)
#define lv(i,a,n) for(int i = a;i >= n;i--)
#define clean(a) memset(a,0,sizeof(a))
const int INF = 1 << 30;
typedef long long ll;
typedef double db;
template <class T>
void read(T &x)
{
    char c;
    bool op = 0;
    while(c = getchar(), c < '0' || c > '9')
        if(c == '-') op = 1;
    x = c - '0';
    while(c = getchar(), c >= '0' && c <= '9')
        x = x * 10 + c - '0';
    if(op) x = -x;
}
template <class T>
void write(T x)
{
    if(x < 0) putchar('-'), x = -x;
    if(x >= 10) write(x / 10);
    putchar('0' + x % 10);
}
struct node
{
    int l,r,nxt;
}a[300010];
int lst[200010],len = 1,cnt = 0;
void add(int x,int y)
{
    a[++len].l = x;
    a[len].r = y;
    a[len].nxt = lst[x];
    lst[x] = len;
}
int n,m,low[200010],dfn[200010];
int from[200010],col[200010];
int du[200010];
int vis[200010];
int top = 0,st[100010],tot = 0;
void tarjan(int u,int g)
{
    bool flag = 0;
    vis[u] = 1;st[++top] = u;
    low[u] = dfn[u] = ++cnt;
    for(int k = lst[u];k;k = a[k].nxt)
    {
        int y = a[k].r;
        if(y == g && !flag)
        {
            flag = 1;
            continue;
        }
        if(!dfn[y])
        {
            tarjan(y,u);
            low[u] = min(low[u],low[y]);
        }
        else if(vis[y])
        low[u] = min(low[u],dfn[y]);
    }
    if(low[u] == dfn[u])
    {
        int p = 0;
        tot++;
        do
        {
            p = st[top--];
            col[p] = tot;
            vis[p] = 0;
        }
        while(p != u);
    }
}
int main()
{
    read(n);read(m);
    duke(i,1,m)
    {
        int x,y;
        read(x);read(y);
        add(x,y);
        add(y,x);
    }
    duke(i,1,n)
        if(!dfn[i])
            tarjan(i,0);
    int ans = 0;
    duke(i,1,n)
    {
        for(int k = lst[i];k;k = a[k].nxt)
        {
            int x = a[k].l;
            int y = a[k].r;
            if(col[x] != col[y])
            {
                du[col[x]]++;
                du[col[y]]++;
            }
        }
    }
//    for(int i = 1; i <= tot; ++i) printf("%d ", du[i]); puts("")
    duke(i,1,tot)
    if(du[i] == 2)
    ans++;
    if(tot == 1)
    printf("0\n");
    else
    printf("%d\n",(ans + 1) / 2);
    return 0;
}