（beginer）DFS (2-SAT) UVA 11294 Wedding

Problem E: Wedding

Up to thirty couples will attend a wedding feast, at which they will be seated on either side of a long table. The bride and groom sit at one end, opposite each other, and the bride wears an elaborate headdress that keeps her from seeing people on the same side as her. It is considered bad luck to have a husband and wife seated on the same side of the table. Additionally, there are several pairs of people conducting adulterous relationships (both different-sex and same-sex relationships are possible), and it is bad luck for the bride to see both members of such a pair. Your job is to arrange people at the table so as to avoid any bad luck.

The input consists of a number of test cases, followed by a line containing 0 0. Each test case gives n, the number of couples, followed by the number of adulterous pairs, followed by the pairs, in the form "4h 2w" (husband from couple 4, wife from couple 2), or "10w 4w", or "3h 1h". Couples are numbered from 0 to n-1 with the bride and groom being 0w and 0h. For each case, output a single line containing a list of the people that should be seated on the same side as the bride. If there are several solutions, any one will do. If there is no solution, output a line containing "bad luck".

Sample Input

10 6
3h 7h
5w 3w
7h 6w
8w 3w
7h 3w
2w 5h
0 0

Possible Output for Sample Input

1h 2h 3w 4h 5h 6h 7h 8h 9h

题意：最多30组人要参加婚礼，然后要分开坐在桌子两边，新郎和新娘在桌子两侧，新娘只能看到桌子对面的人，然后他不能看到妻子和丈夫坐在同一侧，有些人关系也有问题，也不能看到他们坐在同一侧。要怎么安排座位呢？

思路：2-SAT问题，对于一组中的h，要么h坐在对面，要么w坐在对面，把两个作为两个点，对于第i组:2*i表示w坐在对面，2*i+1表示h坐在对面。然后要注意新郎可能也有问题，所以一开始dfs的时候应该先dfs(1)，如果直接就不行，就是不行了，然后再从2开始dfs就行了。最后输出的时候要反过来输出：如果mark[2*i] =true ，输出h ，如果不是就输出w。最后说一下输入的时候用scanf("%d%c")因为可能会输入3h3w 没有空格。。。。人家说的

代码：

#include<iostream>
#include<cstring>
#include<string.h>
#include<cstdio>
#include<vector>
using namespace std;
const int maxn = 1200;
vector<int> G[maxn];
bool mark[maxn];
int S[maxn] , c;
char buffer;
int n , m;

void init()
{
	for (int i = 0 ; i < maxn ; ++i) G[i].clear();
	memset(mark,0,sizeof(mark));
}

void add(int x,int xval,int y,int yval)
{
	x = 2*x+xval;
	y = 2*y+yval;
	G[x].push_back(y^1);
	G[y].push_back(x^1);
}

bool dfs(int x)
{
	if (mark[x^1]) return false;
	if (mark[x]) return true;
	mark[x] = true;
	S[c++] = x;
	for (int i = 0 ; i < G[x].size() ; ++i)
		if (!dfs(G[x][i])) return false; 
	return true;
}

bool Ok()
{
	if (!dfs(1)) return false; 
	for (int i = 2 ; i < 2*n ; i+=2)
	{
		c = 0;
		if (!mark[i] && !mark[i+1])
			if (!dfs(i)) 
			{
				while (c) { mark[S[--c]] = false; }
				if (!dfs(i+1)) return false;
			}
	}
	return true;
}

void input()
{
	while (m--)
	{
		int x , y , xval = 1 , yval = 1;
		scanf("%d%c",&x,&buffer);
		if (buffer=='w') xval = 0;
		scanf("%d%c",&y,&buffer);
		if (buffer=='w') yval = 0;
		add(x,xval,y,yval);
	}
}

void solve()
{
	if (Ok()) {
		bool first = true;
		for (int i = 1 ; i < n ; ++i) 
		{
			if (first) first = false;
			else printf(" ");
			if (mark[2*i]) printf("%dh",i);
			else printf("%dw",i);
		}
	} else { printf("bad luck"); }
	cout << endl;
}

int main()
{
	while (scanf("%d%d",&n,&m),n+m)
	{
		init();
		input();
		solve();
	}
}