题意
给出一张无向图(连通数可能有多个)
1、求出桥有多少个
2、求出有多少边在两个或两个以上的回路中共同存在,求出这些边所在环的边数之和
思路
点双连通分量,tarjan算法,顺便把桥也计数
问题二:找回路,用点双,点双有两种状态,单边和或者环的组合体,判断是否在环中是否有重边,只需要判断没有点双中边的数量和点的数量
1、边数=点数-1 单边
2、边数=点数 单独成环
3、边数>点数 点双为环的组合,说明点双存在几个环重合
如图所示:
所以最终结论就是第三中情况的点双之和。
代码
#include<bits/stdc++.h>
using namespace std;
#define maxn 10005
#define maxm 200005
#define INF 1000000005
#define ll long long
#define IOS ios::sync_with_stdio(false)
struct edge
{
int next, to;
} G[maxm];
int head[maxn], num;
void add(int from, int to)
{
G[++num].next = head[from];
G[num].to = to;
head[from] = num;
}
struct Edge
{
int st, en;
Edge() {}
Edge(int a, int b)
{
st = a, en = b;
}
};
stack <Edge> palm; //保存路径
vector <Edge> block[maxn]; //同一个点双的边保存在一起
int dfn[maxn], low[maxn];
int ind, T, sum1, sum2; //ind为点双的数量
void tarjan(int u, int pre)
{
dfn[u] = low[u] = T++;
for (int i = head[u]; i != -1; i = G[i].next)
{
int v = G[i].to;
if (!dfn[v])
{
palm.push(Edge(u, v));
tarjan(v, u);
if (low[u]>low[v]) low[u] = low[v];
if (dfn[u] <= low[v])
{
for (Edge temp; !palm.empty(); )
{
temp = palm.top();
if (dfn[temp.st]<dfn[v]) break;
block[ind].push_back(temp), palm.pop();
}
block[ind++].push_back(Edge(u, v));
palm.pop();
if (dfn[u]<low[v])
sum1++;
}
}
else if (v != pre && dfn[v]<dfn[u])
{
palm.push(Edge(u, v));
if (low[u]>dfn[v]) low[u] = dfn[v];
}
}
}
bool vis[maxn];
void init()
{
num = 0;
ind = 0, T = 0, sum1 = 0, sum2 = 0;
memset(head, -1, sizeof(head));
memset(dfn, 0, sizeof(dfn));
memset(low, 0, sizeof(low));
memset(vis, 0, sizeof(vis));
}
int main()
{
IOS;
int n, m;
while (cin >> n >> m)
{
if (n == 0 && m == 0)
break;
init();
for (int i = 0; i<m; i++)
{
int x, y;
cin >> x >> y;
add(x, y);
add(y, x);
}
for (int i = 0; i<n; i++)
if (dfn[i] == 0)
tarjan(i, -1);
for (int i = 0; i<ind; i++)
{
int tot = 0;
for (int i = 0; i<n; i++)
vis[i] = false;
for (int j = 0; j<block[i].size(); j++)
{
int u = block[i][j].st, v = block[i][j].en;
if (vis[u] == false)
tot++, vis[u] = true;
if (vis[v] == false)
tot++, vis[v] = true;
}
if (tot<block[i].size())
sum2 += block[i].size();
block[i].clear();
}
cout << sum1 << " " << sum2 << "\n";
}
return 0;
}