hdu.Railway，点双联通分量 + 桥（割边），tarjan算法

Problem - 3394 (hdu.edu.cn)

Railway

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6294    Accepted Submission(s): 2055

 

Problem Description

There are some locations in a park, and some of them are connected by roads. The park manger needs to build some railways along the roads, and he would like to arrange tourist routes to each circuit. If a railway belongs to more than one tourist routes, there might be clash on it, and if a railway belongs to none tourist route, it doesn’t need to build.
Now we know the plan, and can you tell us how many railways are no need to build and how many railways where clash might happen.

 

Input

The Input consists of multiple test cases. The first line of each test case contains two integers, n (0 < n <= 10000), m (0 <= m <= 100000), which are the number of locations and the number of the railways. The next m lines, each line contains two integers, u, v (0 <= u, v < n), which means the manger plans to build a railway on the road between u and v.
You can assume that there is no loop and no multiple edges.
The last test case is followed by two zeros on a single line, which means the end of the input.

 

Output

Output the number of railways that are no need to build, and the number of railways where clash might happen. Please follow the format as the sample.

 

Sample Input

8 10

0 1

1 2

2 3

3 0

3 4

4 5

5 6

6 7

7 4

5 7

0 0

 

Sample Output

1 5

 

Author

momodi@whu

 

题解

对于一个点双联通子图，如果边的个数等于点的个数，那么该点双联通子图刚好形成一个环；如果边的个数大于点的个数，那么该点双联通子图至少存在三个环，且每一条边都至少存在于两个环中，所以该子图的所有边都为冲突边

桥为多余的边

注意：这道题中会有重边

#include<iostream>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<math.h>
#include<map>
using namespace std;
typedef long long LL;
const int N = 1e4 + 5, M = 2e5 + 5;
int n, m;
int h[N], e[M], ne[M], idx;
int dfn[N], low[N], timestamp;
int id[N], siz[N],  dcc_cnt;
int ans1,ans;
stack<pair<int,int>>st;
set<int>pos[N];


void add(int a, int b) {
	e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

void tarjan(int u,int fa) {
	dfn[u] = low[u] = ++timestamp;
	for (int i = h[u]; i != -1; i = ne[i]) {
		int j = e[i];
		if (!dfn[j]) {
			st.push({u,j});
			tarjan(j, i);
			low[u] = min(low[u], low[j]);
			if (low[j] >= dfn[u]) {
				dcc_cnt++;
				id[u] = dcc_cnt;
				pos[dcc_cnt].clear();
				while (1)
				{
					auto t = st.top();
					st.pop();
					siz[dcc_cnt]++;//存某个连通图中的边数
					pos[dcc_cnt].insert(t.first), pos[dcc_cnt].insert(t.second);//存编号为i的连通图中包含的点数
					if (u == t.first && j == t.second)break;
				}
			}
			if (dfn[u] <low[j]) {
				ans1++;
			}
		}
		else if (i != (fa ^ 1) && dfn[j] < dfn[u]) {
			st.push({ u,j });
			low[u] = min(low[u], dfn[j]);
		}
	}
}

int main() {
	while (cin >> n >> m,n||m) {
		memset(h, -1, sizeof h);
		memset(dfn, 0, sizeof dfn);
		memset(siz, 0, sizeof siz);
		idx = timestamp = dcc_cnt = ans1 = ans = 0;
		for (int i = 1,a,b; i <= m; i++) {
			scanf("%d%d", &a, &b);
			add(a, b), add(b, a);
		}
		for (int i = 0; i < n; i++) {
			if (!dfn[i])
				tarjan(i,-1);
		}
		for (int i = 1; i <= dcc_cnt; i++) {
			if (siz[i] > pos[i].size()) {
				ans += siz[i];
			}
		}
		cout << ans1 << " " << ans << endl;
	}
	return 0;
}