21 | 1月 | 2016 | 超客的水墨烟雨

poj2942

maksyuki 发表于 oj 分类，标签: 图论-求解割点\割边(Tarjan)

21 1月 2016

Knights of the Round Table

Being a knight is a very attractive career: searching for the Holy Grail, saving damsels in distress, and drinking with the other knights are fun things to do. Therefore, it is not very surprising that in recent years the kingdom of King Arthur has experienced an unprecedented increase in the number of knights. There are so many knights now, that it is very rare that every Knight of the Round Table can come at the same time to Camelot and sit around the round table; usually only a small group of the knights isthere, while the rest are busy doing heroic deeds around the country.

Knights can easily get over-excited during discussions-especially after a couple of drinks. After some unfortunate accidents, King Arthur asked the famous wizard Merlin to make sure that in the future no fights break out between the knights. After studying the problem carefully, Merlin realized that the fights can only be prevented if the knights are seated according to the following two rules:

The knights should be seated such that two knights who hate each other should not be neighbors at the table. (Merlin has a list that says who hates whom.) The knights are sitting around a roundtable, thus every knight has exactly two neighbors.
An odd number of knights should sit around the table. This ensures that if the knights cannot agree on something, then they can settle the issue by voting. (If the number of knights is even, then itcan happen that yes" and no" have the same number of votes, and the argument goes on.)

Merlin will let the knights sit down only if these two rules are satisfied, otherwise he cancels the meeting. (If only one knight shows up, then the meeting is canceled as well, as one person cannot sit around a table.) Merlin realized that this means that there can be knights who cannot be part of any seating arrangements that respect these rules, and these knights will never be able to sit at the Round Table (one such case is if a knight hates every other knight, but there are many other possible reasons). If a knight cannot sit at the Round Table, then he cannot be a member of the Knights of the Round Table and must be expelled from the order. These knights have to be transferred to a less-prestigious order, such as the Knights of the Square Table, the Knights of the Octagonal Table, or the Knights of the Banana-Shaped Table. To help Merlin, you have to write a program that will determine the number of knights that must be expelled.

Input

The input contains several blocks of test cases. Each case begins with a line containing two integers 1 ≤ n ≤ 1000 and 1 ≤ m ≤ 1000000 . The number n is the number of knights. The next m lines describe which knight hates which knight. Each of these m lines contains two integers k1 and k2 , which means that knight number k1 and knight number k2 hate each other (the numbers k1 and k2 are between 1 and n ).

The input is terminated by a block with n = m = 0 .

Output

For each test case you have to output a single integer on a separate line: the number of knights that have to be expelled.

Sample Input

5 5
1 4
1 5
2 5
3 4
4 5
0 0

Sample Output

Hint

Huge input file, 'scanf' recommended to avoid TLE.

Source

Central Europe 2005

题目类型：反向构图+求解重连通分量(点)+交叉染色判奇圈

算法分析：首先构造图，把武士看作是顶点，若两个武士之间没有相互仇视，则在两个顶点之间连边，表示两个武士可以在一个圆桌旁相邻就坐。要让武士在一个圈内才能满足要求，不在圈内的武士会被剔除掉。所以这个时候求解这个图的重连通分量(点)，构造满足条件的环。此时只要判断这个环是奇圈即可，这里使用交叉染色的方法判定(由一个判奇圈的充要条件得来)。注意：构造出来的图有可能是非连通图！！！

/*************************************************
Author        :supermaker
Created Time  :2016/1/21 18:40:28
File Location :C:\Users\abcd\Desktop\TheEternalPoet
**************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)
#define CPPFF ifstream cin ("aaa.txt")
#define	DB(ccc)	cout << #ccc << " = " << ccc << endl
#define PB push_back
#define MP(A, B) make_pair(A, B)

typedef long long LL;
typedef unsigned long long ULL;
typedef double DB;
typedef pair <int, int> PII;
typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 1000 + 66;

int g[maxn][maxn], vis[maxn], dfsn[maxn], low[maxn], color[maxn], part[maxn], res[maxn];
int n, m, id;
stack <PII> sta;

void Init (int rt)
{
	vis[rt] = 1;
	dfsn[rt] = low[rt] = id = 1;
	while (!sta.empty ()) sta.pop ();
}

bool CrossColor (int u, int col)
{
	color[u] = col;
	for (int v = 1; v <= n; v++)
		if (g[u][v] && part[v])//it is possible : g[u][v] == 2 || g[u][v] == 1
		{
			if (color[v] == 0 && CrossColor (v, -col)) return 1;
			if (color[v] == col) return 1;
		}
	return 0;
}

void SS ()
{
	for (int i = 1; i <= n; i++)
		if (part[i])
		{
			memset (color, 0, sizeof (color));
			if (CrossColor (i, 1))
				for (int j = 1; j <= n; j++)
					if (part[j])
						res[j]++;
			return ;
		}
}

void dfs (int u, int par)
{
	for (int v = 1; v <= n; v++)
	{
		if (g[u][v] == 1)
		{
			sta.push (PII (u, v));
			g[u][v] = g[v][u] = 2;
			if (!vis[v])
			{
				vis[v] = 1;
				dfsn[v] = low[v] = ++id;
				dfs (v, u);
				low[u] = min (low[u], low[v]);
				if (low[v] >= dfsn[u])
				{
					memset (part, 0, sizeof (part));
					while (!sta.empty ())
					{
						PII temp = sta.top (); sta.pop ();
						part[temp.first] = part[temp.second] = 1;	
						//printf (" %d-%d ", temp.first, temp.second);
						if ((temp.first == u && temp.second == v) || (temp.first == v && temp.second == u))
							break;	
					}
					//puts ("");
					SS ();
				}
			}
			else if (v != par)
				low[u] = min (low[u], dfsn[v]);
		}
	}
}

int main()
{
    //CFF;
    //CPPFF;
	while (scanf ("%d%d", &n, &m) != EOF)
	{
		if (n == 0 && m == 0) break;

		memset (res, 0, sizeof (res));
		memset (vis, 0, sizeof (vis));
		for (int i = 0; i < maxn; i++)
			for (int j = 0; j < maxn; j++)
			{
				if (i == j) g[i][j] = 0;//important!!!
				else g[i][j] = 1;
			}

		for (int i = 1; i <= m; i++)
		{
			int u, v;
			scanf ("%d%d", &u, &v);
			g[u][v] = g[v][u] = 0;
		}

		for (int i = 1; i <= n; i++)
			if (!vis[i])
			{
				Init (i);
				dfs (i, -1);
			}
		int ans = 0;
		for (int i = 1; i <= n; i++)
			if (!res[i]) 
				ans++;

		printf ("%d\n", ans);
	}
    return 0;
}

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/*************************************************

Author :supermaker

Created Time :2016/1/21 18:40:28

File Location :C:\Users\abcd\Desktop\TheEternalPoet

**************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)

#define CPPFF ifstream cin ("aaa.txt")

#define DB(ccc) cout << #ccc << " = " << ccc << endl

#define PB push_back

#define MP(A, B) make_pair(A, B)

typedef long long LL;

typedef unsigned long long ULL;

typedef double DB;

typedef pair <int, int> PII;

typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 1000 + 66;

int g[maxn][maxn], vis[maxn], dfsn[maxn], low[maxn], color[maxn], part[maxn], res[maxn];

int n, m, id;

stack <PII> sta;

void Init (int rt)

{

vis[rt] = 1;

dfsn[rt] = low[rt] = id = 1;

while (!sta.empty ()) sta.pop ();

}

bool CrossColor (int u, int col)

{

color[u] = col;

for (int v = 1; v <= n; v++)

if (g[u][v] && part[v])//it is possible : g[u][v] == 2 || g[u][v] == 1

{

if (color[v] == 0 && CrossColor (v, -col)) return 1;

if (color[v] == col) return 1;

}

return 0;

}

void SS ()

{

for (int i = 1; i <= n; i++)

if (part[i])

{

memset (color, 0, sizeof (color));

if (CrossColor (i, 1))

for (int j = 1; j <= n; j++)

if (part[j])

res[j]++;

return ;

}

void dfs (int u, int par)

{

for (int v = 1; v <= n; v++)

{

if (g[u][v] == 1)

{

sta.push (PII (u, v));

g[u][v] = g[v][u] = 2;

if (!vis[v])

{

vis[v] = 1;

dfsn[v] = low[v] = ++id;

dfs (v, u);

low[u] = min (low[u], low[v]);

if (low[v] >= dfsn[u])

{

memset (part, 0, sizeof (part));

while (!sta.empty ())

{

PII temp = sta.top (); sta.pop ();

part[temp.first] = part[temp.second] = 1;

//printf (" %d-%d ", temp.first, temp.second);

if ((temp.first == u && temp.second == v) || (temp.first == v && temp.second == u))

break;

}

//puts ("");

SS ();

}

else if (v != par)

low[u] = min (low[u], dfsn[v]);

}

int main()

{

//CFF;

//CPPFF;

while (scanf ("%d%d", &n, &m) != EOF)

{

if (n == 0 && m == 0) break;

memset (res, 0, sizeof (res));

memset (vis, 0, sizeof (vis));

for (int i = 0; i < maxn; i++)

for (int j = 0; j < maxn; j++)

{

if (i == j) g[i][j] = 0;//important!!!

else g[i][j] = 1;

}

for (int i = 1; i <= m; i++)

{

int u, v;

scanf ("%d%d", &u, &v);

g[u][v] = g[v][u] = 0;

}

for (int i = 1; i <= n; i++)

if (!vis[i])

{

Init (i);

dfs (i, -1);

}

int ans = 0;

for (int i = 1; i <= n; i++)

if (!res[i])

ans++;

printf ("%d\n", ans);

}

return 0;

}

zoj2588

maksyuki 发表于 oj 分类，标签: 图论-求解割点\割边(Tarjan)

21 1月 2016

Burning Bridges

Ferry Kingdom is a nice little country located on N islands that are connected by M bridges. All bridges are very beautiful and are loved by everyone in the kingdom. Of course, the system of bridges is designed in such a way that one can get from any island to any other one.

But recently the great sorrow has come to the kingdom. Ferry Kingdom was conquered by the armies of the great warrior Jordan and he has decided to burn all the bridges that connected the islands. This was a very cruel decision, but the wizards of Jordan have advised him no to do so, because after that his own armies would not be able to get from one island to another. So Jordan decided to burn as many bridges as possible so that is was still possible for his armies to get from any island to any other one.

Now the poor people of Ferry Kingdom wonder what bridges will be burned. Of course, they cannot learn that, because the list of bridges to be burned is kept in great secret. However, one old man said that you can help them to find the set of bridges that certainly will not be burned.

So they came to you and asked for help. Can you do that?

Input

The input contains multiple test cases. The first line of the input is a single integer T (1 <= T <= 20) which is the number of test cases. T test cases follow, each preceded by a single blank line.

The first line of each case contains N and M - the number of islands and bridges in Ferry Kingdom respectively (2 <= N <= 10 000, 1 <= M <= 100 000). Next M lines contain two different integer numbers each and describe bridges. Note that there can be several bridges between a pair of islands.

Output

On the first line of each case print K - the number of bridges that will certainly not be burned. On the second line print K integers - the numbers of these bridges. Bridges are numbered starting from one, as they are given in the input.

Two consecutive cases should be separated by a single blank line. No blank line should be produced after the last test case.

Sample Input

6 7

1 2

2 3

2 4

5 4

1 3

4 5

3 6

10 16

2 6

3 7

6 5

5 9

5 4

1 2

9 8

6 4

2 10

3 8

7 9

1 4

2 4

10 5

1 6

6 10

Sample Output

3 7

Author: Andrew Stankevich

Source: Andrew Stankevich's Contest #5

题目类型：求解割边(可能有重边) Tarjan

算法分析：求解割边的过程和求解割点的类似，只是满足：low[v] > dfsn[u]，u是v的父节点。若图中存在重边，则直接特判即可，注意输出的格式

/*************************************************
Author        :supermaker
Created Time  :2016/1/21 13:05:23
File Location :C:\Users\abcd\Desktop\TheEternalPoet
**************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")
#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>
using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)
#define CPPFF ifstream cin ("aaa.txt")
#define	DB(ccc)	cout << #ccc << " = " << ccc << endl
#define PB push_back
#define MP(A, B) make_pair(A, B)

typedef long long LL;
typedef unsigned long long ULL;
typedef double DB;
typedef pair <int, int> PII;
typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int maxn = 10000 + 66;

int vis[maxn], dfsn[maxn], low[maxn], out[maxn*10];
int n, m, id, len;
map <PII, int> overedge;
vector <PII> edge[maxn*10];

void Init (int rt)
{
	overedge.clear ();
	for (int i = 0; i < maxn * 10; i++)
		edge[i].clear ();
	dfsn[rt] = low[rt] = id = 1;
	memset (vis, 0, sizeof (vis));
	vis[rt] = 1;
	len = 0;
}

void dfs (int u, int par)
{
	for (int i = 0; i < edge[u].size (); i++)
	{
		int v = edge[u][i].first;
		if (!vis[v])
		{
			vis[v] = 1;
			dfsn[v] = low[v] = ++id;
			dfs (v, u);
			low[u] = min (low[u], low[v]);
			if (low[v] > dfsn[u] && overedge[PII(u,v)] == 1)
				out[++len] = edge[u][i].second;
		}

		else if (v != par)
			low[u] = min (low[u], dfsn[v]);
	}
}

int main()
{
    //CFF;
    //CPPFF;
	int t, flag = 1;
	scanf ("%d", &t);
	while (t--)
	{
		if (flag++ > 1) puts ("");
		scanf ("%d%d", &n, &m);

		Init (1);
		for (int i = 1; i <= m; i++)
		{
			int u, v;
			scanf ("%d%d", &u, &v);
			edge[u].push_back (PII (v, i));
			edge[v].push_back (PII (u, i));
			if (overedge.find (PII (u, v)) == overedge.end ())
				overedge[PII(u,v)] = overedge[PII(v,u)] = 1;

			else
				overedge[PII(u,v)]++, overedge[PII(v,u)]++;
		}

		dfs (1, -1);

		printf ("%d\n", len);
		sort (out + 1, out + 1 + len);
		for (int i = 1; i <= len; i++)
		{
			if (i == 1) printf ("%d", out[i]);
			else printf (" %d", out[i]);
		}
		if (len) puts ("");
	}
    return 0;
}

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/*************************************************

Author :supermaker

Created Time :2016/1/21 13:05:23

File Location :C:\Users\abcd\Desktop\TheEternalPoet

**************************************************/

#pragma comment(linker, "/STACK:102400000,102400000")

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

using namespace std;

#define CFF freopen ("aaa.txt", "r", stdin)

#define CPPFF ifstream cin ("aaa.txt")

#define DB(ccc) cout << #ccc << " = " << ccc << endl

#define PB push_back

#define MP(A, B) make_pair(A, B)

typedef long long LL;

typedef unsigned long long ULL;

typedef double DB;

typedef pair <int, int> PII;

typedef pair <int, bool> PIB;

const int INF = 0x7F7F7F7F;

const int MOD = 1e9 + 7;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int maxn = 10000 + 66;

int vis[maxn], dfsn[maxn], low[maxn], out[maxn*10];

int n, m, id, len;

map <PII, int> overedge;

vector <PII> edge[maxn*10];

void Init (int rt)

{

overedge.clear ();

for (int i = 0; i < maxn * 10; i++)

edge[i].clear ();

dfsn[rt] = low[rt] = id = 1;

memset (vis, 0, sizeof (vis));

vis[rt] = 1;

len = 0;

}

void dfs (int u, int par)

{

for (int i = 0; i < edge[u].size (); i++)

{

int v = edge[u][i].first;

if (!vis[v])

{

vis[v] = 1;

dfsn[v] = low[v] = ++id;

dfs (v, u);

low[u] = min (low[u], low[v]);

if (low[v] > dfsn[u] && overedge[PII(u,v)] == 1)

out[++len] = edge[u][i].second;

}

else if (v != par)

low[u] = min (low[u], dfsn[v]);

}

int main()

{

//CFF;

//CPPFF;

int t, flag = 1;

scanf ("%d", &t);

while (t--)

{

if (flag++ > 1) puts ("");

scanf ("%d%d", &n, &m);

Init (1);

for (int i = 1; i <= m; i++)

{

int u, v;

scanf ("%d%d", &u, &v);

edge[u].push_back (PII (v, i));

edge[v].push_back (PII (u, i));

if (overedge.find (PII (u, v)) == overedge.end ())

overedge[PII(u,v)] = overedge[PII(v,u)] = 1;

else

overedge[PII(u,v)]++, overedge[PII(v,u)]++;

}

dfs (1, -1);

printf ("%d\n", len);

sort (out + 1, out + 1 + len);

for (int i = 1; i <= len; i++)

{

if (i == 1) printf ("%d", out[i]);

else printf (" %d", out[i]);

}

if (len) puts ("");

}

return 0;

}

超客的水墨烟雨

poj2942

zoj2588

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