DFS相关的一些模板

/*请从main开始看*/

/*
SampleInput:

6 8 7
0 5
2 4
2 3
1 2
0 1
3 4
3 5
0 2
1 4
0 0
1 1
2 2
3 3
4 4
5 5
4 2 2
0 1
2 3
2 1
0 0

*/

#include<cstdio>
#include<cstring>
#include<list>
#include<stack>
#include<set>
using namespace std;
const int maxn = 100002;

list<int> point[maxn];///当然，你也可以使用vector实现
set<int> cc[maxn];///cc指连通分量
bool vis[maxn], color[maxn];
int edgeTo[maxn];///dfs出的路径
stack<int> path;
int beg, end, count;
bool isend, hascycle, isTwoColorable;

///复杂度：O(V+E)
///找路，从v到end
void dfs(int v)
{
	vis[v] = true;
	list<int>::iterator it;
	for (it = point[v].begin(); it != point[v].end(); ++it)
	{
		if (isend)
			return;
		int w = (int) * it;
		if (!vis[w])
		{
			edgeTo[w] = v;
			if (w == end)
			{
				isend = true;
				return;
			}
			dfs(w);
		}
	}
}

///找环，从beg到beg
void dfs(int v, int u)
{
	vis[v] = true;
	list<int>::iterator it;
	for (it = point[v].begin(); it != point[v].end(); ++it)
	{
		if (hascycle)
			return;
		int w = (int) * it;
		if (!vis[w])
		{
			edgeTo[w] = v;
			dfs(w, v);
		}
		else if (w != u && w == beg)
			///该已找到的点不是前一个点(上一层dfs的点)
			///说明从该点出发可以沿着另一条路回来，所以G有环。
		{
			edgeTo[w] = v;
			hascycle = true;
			return;
		}
	}
}

///简单找环
void simpledfs(int v, int u)
{
	vis[v] = true;
	list<int>::iterator it;
	for (it = point[v].begin(); it != point[v].end(); ++it)
	{
		if (hascycle)
			return;
		int w = (int) * it;
		if (!vis[w])
			dfs(w, v);
		else if (w != u)
			///该已找到的点不是前一个点(上一层dfs的点)
			///说明从该点出发可以沿着另一条路回来，所以G有环。
		{
			hascycle = true;
			return;
		}
	}
}

///双色否
void colordfs(int v)
{
	vis[v] = true;
	list<int>::iterator it;
	for (it = point[v].begin(); it != point[v].end(); ++it)
	{
		if (!isTwoColorable)
			return;
		int w = (int) * it;
		if (!vis[w])
		{
			color[w] = !color[v];
			colordfs(w);
		}
		else if (color[w] == color[v])
		{
			isTwoColorable = false;
			return;
		}
	}
}

///找连通分量，遍历型
void exdfs(int v)
{
	vis[v] = true;
	cc[count].insert(v);
	list<int>::iterator it;
	for (it = point[v].begin(); it != point[v].end(); ++it)
	{
		int w = (int) * it;
		if (!vis[w])
			exdfs(w);
	}
}

///应用1
///输出起点v到终点w的路径，以空格隔开
///如果不存在，则输出-1
inline void printpath(int v, int w)
{
	memset(vis, 0, sizeof(vis));
	memset(edgeTo, -1, sizeof(edgeTo));
	isend = false;
	while (!path.empty())
		path.pop();
	if (v != w)
	{
		end = w;
		dfs(v);
		for (int x = w; x != v; x = edgeTo[x])///先从w到v
		{
			if (~edgeTo[x]) /// edgeTo[x] != -1
				path.push(x);
			else ///说明v,w在两个不同的连通分量上
			{
				printf("-1\n");
				return;
			}
		}
		printf("%d", v);
		while (!path.empty())
		{
			printf(" %d", path.top());
			path.pop();
		}
		printf("\n");
	}
	else///找一个环
	{
		hascycle = false;
		beg = v;
		dfs(v, v);
		if (hascycle)
		{
			for (int x = edgeTo[v] ; x != v; x = edgeTo[x])
				path.push(x);
			printf("%d", v);
			while (!path.empty())
			{
				printf(" %d", path.top());
				path.pop();
			}
			printf(" %d\n", v);
		}
		else
			printf("-1\n");
	}
}

///应用2
///判断是否为无环图（假设不存在自环或平行边）
inline void hasCycle(int n)
{
	memset(vis, 0, sizeof(vis));
	hascycle = false;
	for (int i = 0; i < n; ++i)
		if (!vis[i])
			simpledfs(i, i);
}

///应用3
///返回连通子图的个数
inline int count_connected_subgraph(int n)
{
	memset(vis, 0, sizeof(vis));
	for (int i = 1; i <= count; ++i)
		cc[i].clear();
	count = 1;
	for (int i = 0; i < n; ++i)
	{
		if (!vis[i])
		{
			exdfs(i);
			++count;
		}
	}
	return count - 1;
}

///应用4
///判断G是否为二分图
inline void checkTwoColor(int n)
{
    memset(vis, 0, sizeof(vis));
	memset(color, 0, sizeof(color));
	isTwoColorable = true;
	for (int i = 0; i < n; ++i)
		if (!vis[i])
			colordfs(i);
}

int main(void)
{
	int n, m, k, temp, v, w;///点数=n，边数=m，查询数=k
	while (~scanf("%d%d%d", &n, &m, &k))
	{
		///0. 初始化
		temp = m;
		while (temp--)
		{
			scanf("%d%d", &v, &w);///这里输入要求为0~(n-1)
			point[v].push_front(w);
			point[w].push_front(v);
		}

		///1. 判断图G是否有环
		hasCycle(n);
		printf(hascycle ? "YES\n" : "NO\n");

		///2. 判断图G是否为二分图
		checkTwoColor(n);
		printf(isTwoColorable ? "YES\n" : "NO\n");

		///3. 连通子图的个数及每个连通子图所包含的点(升序输出)
		printf("%d\n", count_connected_subgraph(n));
		for (int i = 1; i < count; ++i)
		{
			set<int>::iterator it;
			for (it = cc[i].begin(); it != cc[i].end(); ++it)
				printf("%d ", *it);
			printf("\n");
		}

		///4. 判断是否存在v到w的路径/求v到w的路径(保证v,w在范围内)
		while (k--)
		{
			scanf("%d%d", &v, &w);
			printpath(v, w);
		}

		///0. 初始化
		for (int i = 0; i < n; ++i)
			point[i].clear();
		printf("\n");
	}
	return 0;
}

待实现：欧拉回路