782E. Underground Lab DFS 好题

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题意:给出一个图,有n个点,m条边,k个人,每个人至多只能走$lceilfrac{2n}{k} ceil$步,输出可行的方案即输出每个人所走的步数和所走点

思路: 由于保证给出的是连通图，且每人$lceilfrac{2n}{k} ceil$的步数都用完的话，显然必定有答案。那么我们不妨构造一个DFS的遍历序列（包括回溯的过程）,那么最后对这个序列进行分配即可,如果图已经走完,而还有人没走,那么直接输出 1 1 即可

/** @Date    : 2017-05-10 18:06:33
  * @FileName: 782E DFS.cpp
  * @Platform: Windows
  * @Author  : Lweleth (SoundEarlf@gmail.com)
  * @Link    : https://github.com/Lweleth
  * @Version : $Id$
  */
#include <bits/stdc++.h>
#define LL long long
#define PII pair
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 1e5+20;
const double eps = 1e-8;

int n, m, k;
vector vt[2*N];
int a[2*N];
int ans[2*N];
bool vis[N];

int cnt = 0;
void dfs(int x)
{
	vis[x] = 1;
	a[cnt++] = x;
	int l = vt[x].size();
	for(int i = 0; i < l; i++)
	{
		int np = vt[x][i];
		if(vis[np])
			continue;
		dfs(np);
		a[cnt++] = x;//模拟回溯
	}
}

int main()
{
	while(cin >> n >> m >> k)
	{
		for(int i = 0; i < m; i++)
		{
			int x, y;
			scanf("%d%d", &x, &y);
			vt[x].PB(y);
			vt[y].PB(x);
		}
		int ma = 2*n/k + ((2*n)%k?1:0);
		MMF(vis);
		cnt = 0;
		dfs(1);
		int cc = 0;
		for(int i = 0; i < cnt; i++)
		{
			//cout << a[i]<<"~";
			if(cc < ma)
				ans[cc++] = a[i];
			else
			{
				k--;
				printf("%d ", ma);
				for(int j = 0; j < cc; j++)
					printf("%d%s", ans[j], j==cc-1?"
":" ");
				cc = 0;
				ans[cc++] = a[i]; 
			}
		}
		if(cc > 0)
			printf("%d ", cc), k--;
		for(int i = 0; i < cc; i++)
			printf("%d%s", ans[i], i==cc-1?"
":" ");
		for(int i = 0; i < k; i++)
			printf("1 1
");
	}
    return 0;
}