题目名称

华华和月月逛公园

题目大意

给定n个点，m条边，找出其中非割边的数量，即m-割边数量

实现思路

求割边数量，直接Tarjan算法板子即可。
《算法竞赛进阶指南》P397

代码实现

#include<bits/stdc++.h>
using namespace std;
#define inf 0x3f3f3f3f
const double eps=1e-8;
const int maxn=1e5+7;
const int SIZE = 1e6;
int head[SIZE], ver[SIZE * 2], Next[SIZE * 2];
int dfn[SIZE], low[SIZE];
int n, m, tot, num;
bool bridge[SIZE * 2];
void add(int x, int y) {
    ver[++tot] = y, Next[tot] = head[x], head[x] = tot;
}
void tarjan(int x, int in_edge)
{
    dfn[x] = low[x] = ++num;
    for (int i = head[x]; i; i = Next[i])
    {
        int y = ver[i];
        if (!dfn[y]) {
            tarjan(y, i);
            low[x] = min(low[x], low[y]);
            if (low[y] > dfn[x])
                bridge[i] = bridge[i ^ 1] = true;
        }
        else if (i != (in_edge ^ 1))
            low[x] = min(low[x], dfn[y]);
    }
}
int main() 
{
    cin >> n >> m;
    tot = 1;
    for (int i = 1; i <= m; i++) {
        int x, y;
        scanf("%d%d", &x, &y);
        add(x, y), add(y, x);
    }
    for (int i = 1; i <= n; i++)
        if (!dfn[i]) tarjan(i, 0);
    int bridgeCnt=0;
    for (int i = 2; i < tot; i += 2)
        if (bridge[i]){
            //printf("%d %d\n", ver[i ^ 1], ver[i]);
            bridgeCnt++;
        }
    printf("%d\n",m-bridgeCnt);
    return 0;
}