POJ 3177 Redundant Paths（无向图缩点）

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;

const int N=50010;

struct node
{
    int v, next;
} no[N];
int r[N], be[N], Stack[N], vis[N], head[N], dfn[N], low[N];
bool use[5010][5010];
int ans, Time, top, k;

void Init()
{
    memset(head, -1, sizeof(head));
    memset(dfn, 0, sizeof(dfn));
    memset(low, 0, sizeof(low));
    memset(vis, 0, sizeof(vis));
    memset(r, 0, sizeof(r)); ///r数组存放与第i个点相邻的点与其强连通分量不同的个数

    Time = top = ans = k = 0;
}

void Add(int a, int b)
{
    no[k].v = b;
    no[k].next = head[a];

    head[a] = k++;
}

void Tarjan(int u, int fa)
{
    int v, i;

    Stack[top++] = u;
    vis[u] = 1;
    dfn[u] = low[u] = ++Time;

    for (i = head[u]; i != -1; i = no[i].next)
    {
        v = no[i].v;

        if (!dfn[v])
        {
            Tarjan(v, u);
            low[u] = min(low[u], low[v]);
        }
        else if (v != fa) low[u] = min(low[u], dfn[v]);
    }

    if (dfn[u] == low[u])
    {
        ++ans;
        do
        {
            v = Stack[--top];
            vis[v] = 0;
            be[v] = ans;
        }while(u != v);
    }
}

int main ()
{
    int n, m, a, b, i, j, num;

    while (scanf("%d%d", &n, &m) != EOF)
    {
        Init();
        num = 0;

        while (m--)
        {
            scanf("%d%d", &a, &b);

            if (!use[a][b]) ///去重边
            {
                Add(a, b);
                Add(b, a);
                use[a][b] = use[b][a] = 1;
            }
        }

        Tarjan(1, 1); ///因为任意的两个点都至少有一条边，所以查找一次就行

        for (i = 1; i <= n; i++)
        {
            for (j = head[i]; j != -1; j = no[j].next)
            {
                if (be[i] != be[no[j].v]) ///be数组存放的是该点属于第几个强连通分量
                    r[be[i]]++;
            }
        }

        for (i = 1; i <= ans; i++)
            if (r[i] == 1) num++; ///只有当等于1时，该点才是叶子节点

        printf("%d\n", (num+1)/2); ///让每个叶子节点至少连接两条边，只有这样才能让任意两点之间至少有两条路
    }

    return 0;
}