kosarajo求强连通分量

kosarajo求强连通分量的证明

• 因为根据反向图的dfs求出的拓扑序列使得原本的DAG图中点的搜索优先级倒转
• 所以在原图dfs会优先将最末端的点优先跑完，而上面的点再跑时，因为下面的点已经被标记过，阻止了连通块的扩张
• 按照DAG的逆优先顺序，从下到上，依次把强连通分量“挖出来”

例题

poj2186

题目链接

题意

• n个点m条边，求有多少个点是任意点均可达到的

思路

• 根据反向图求出拓扑序列，再使用正向dfs求出连通分量，题目所求即第一个强连通分量中点的个数
• 注意：图本身可能是非联通的，要特判

代码

#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <set>
#include <algorithm>
using namespace std;
const int N = 10005;
vector<int> g[N], gr[N], path;
int vis[N], sscid[N], n, m;
void dfs1(int x) {
	vis[x] = 1;
	for(unsigned int i = 0; i < gr[x].size(); ++ i) {
		int y = gr[x][i];
		if(!vis[y]) {
			dfs1(y);
		}
	}
	path.push_back(x);
	return;
}
void dfs2(int x, int cnt) {
	sscid[x] = cnt;
	for(unsigned int i = 0; i < g[x].size(); ++ i) {
		int y = g[x][i];
		if(!sscid[y]) {
			dfs2(y, cnt);
		}
	}
	return;
}
void dfs3(int x) {
	vis[x] = 1;
	for(unsigned int i = 0; i < gr[x].size(); ++ i) {
		int y = gr[x][i];
		if(!vis[y]) {
			dfs3(y);
		}
	}
}
void kosarajo() {
	int cnt = 0;
	for(int i = 1; i <= n; ++ i) {
		if(!vis[i]) {
			dfs1(i);
		}
	}
	reverse(path.begin(), path.end());
	for(unsigned int i = 0; i < path.size(); ++ i) {
		int x = path[i];
		if(!sscid[x]) {
			dfs2(x, ++ cnt);
		}
	}
}
void init() {
	for(int i = 0; i < N; ++ i) {
		g[i].clear();
		gr[i].clear();
	}
	memset(vis, 0, sizeof(vis));
	memset(sscid, 0, sizeof(sscid));
	return;
}
int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin >> n >> m;
	init();
	for(int i = 1; i <= m; ++ i) {
		int u, v;
		cin >> u >> v;
		g[u].push_back(v);
		gr[v].push_back(u);
	}
	kosarajo();
	int count = 0, x = 0;
	for(int i = 1; i <= n; ++ i) {
		if(sscid[i] == 1) {
			++ count;
			x = i;
		}
	}
	memset(vis, 0, sizeof(vis));
	dfs3(x);
	bool flag = true;
	for(int i = 1; i <= n; ++ i) {
		if(!vis[i]) {
			flag = false;
			break;
		}
	}
	if(!flag) cout << 0 << "\n";
	else cout << count << "\n";
	return 0;
}

hdu1269

题目链接

题意

• 如题

思路

• 用kosarajo求出分量个数看是否为1

代码

#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
const int N = 10005;
int n, m, vis[N], sscid[N];
vector<int> g[N], gr[N], path;
void dfs1(int x) {
	vis[x] = 1;
	for(auto y : gr[x]) {
		if(!vis[y]) {
			dfs1(y);
		}
	}
	path.push_back(x);
	return;
}
void dfs2(int x, int cnt) {
	sscid[x] = cnt;
	for(auto y : g[x]) {
		if(!sscid[y]) {
			dfs2(y, cnt);
		}
	}
	return;
}
int kosarajo() {
	int cnt = 0;
	for(int i = 1; i <= n; ++ i) {
		if(!vis[i]) {
			dfs1(i);
		}
	}
	reverse(path.begin(), path.end());
	for(auto x : path) {
		if(!sscid[x]) {
			dfs2(x, ++ cnt);
		}
	}
	return cnt;
}
void init() {
	memset(vis, 0, sizeof(vis));
	memset(sscid, 0, sizeof(sscid));
	path.clear();
	for(int i = 0; i < N; ++ i) {
		g[i].clear();
		gr[i].clear();
	}
}
void slove() {
	init();
	for(int i = 1; i <= m; ++ i) {
		int u, v;
		cin >> u >> v;
		g[u].push_back(v);
		gr[v].push_back(u);
	}
	if(kosarajo() == 1) printf("Yes\n");
	else printf("No\n");
	return;
}
int main() {
	while(scanf("%d %d", &n, &m), n || m) {
		slove();
	}
	return 0;
}