【POJ1417】【True Liars】【加权并查集+背包+输出路径】

2 1 1
1 2 no
2 1 no
3 2 1
1 1 yes
2 2 yes
3 3 yes
2 2 1
1 2 yes
2 3 no
5 4 3
1 2 yes
1 3 no
4 5 yes
5 6 yes
6 7 no
0 0 0

no
no
1
2
end
3
4
5
6
end

#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <string>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
using namespace std;
    
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define mp push_back

int n,p1,p2;
char str[5];
int F[610];
int val[610];
int cnt;
int a[610][2]; //0: 1:
vector<int> b[610][2];
int dp[610][610]; //用了i个物品，好人数目为j的方案数目
int pre[610][610];
bool used[610];

int Find(int x)
{
	if(F[x] == -1) return x;
	int t = Find(F[x]);
	val[x] ^= val[F[x]];
	return F[x] = t;
}

void merge(int x,int y,int op)
{
	int fx = Find(x);
	int fy = Find(y);
	if(fx != fy)
	{
		F[fx] = fy;
		val[fx] = val[x]^val[y]^op;
	}
}
int main()
{
    while(scanf("%d%d%d",&n,&p1,&p2))
	{
		if(n == 0 && p1 == 0 && p2 == 0) break;
		memset(F,-1,sizeof(F));
		memset(val,0,sizeof(val));
		bool ok = true;
		cnt = 1;
		for(int i=0;i<n;i++)
		{
			int u,v,op;
			scanf("%d%d%s",&u,&v,str);
			if(!ok) continue;
			op = (str[0] == 'y' ? 0 : 1);
			int t1 = Find(u);
			int t2 = Find(v);
			if(t1 == t2)
			{
			}
			else
			{
				merge(u,v,op);
			}
		}
		if(!ok)
		{
			printf("no\n");
			continue;
		}
		
		for(int i=0;i<=p1+p2;i++)
		{
			a[i][0] = 0;
			a[i][1] = 0;
			b[i][0].clear();
			b[i][1].clear();
		}
		
		memset(used,0,sizeof(used));
		for(int i=1;i<=p1+p2;i++)
		{
			if(used[i]) continue;
			int tmp = Find(i);
			for(int j=1;j<=p1+p2;j++)
			{
				if(tmp == Find(j) && !used[j])
				{
					used[j] = true;
					a[cnt][val[j]] ++;
					b[cnt][val[j]].push_back(j);
				}
			}
			cnt ++;
		}
	
		memset(dp,0,sizeof(dp));
		memset(pre,-1,sizeof(pre));
		dp[0][0] = 1;
		for(int i=1;i<cnt;i++)
		{
			for(int j=p1;j>=0;j--)
			{
				if(j >= a[i][0]  && dp[i-1][j-a[i][0]])
				{
					dp[i][j] += dp[i-1][j-a[i][0]];
					pre[i][j] = j-a[i][0];

				}
				if(j >= a[i][1] && dp[i-1][j-a[i][1]])
				{
					dp[i][j] += dp[i-1][j-a[i][1]];
					pre[i][j] = j-a[i][1];
				}
			}
		}
		if(dp[cnt-1][p1] != 1)
		{
			printf("no\n");
		}
		else
		{
			vector<int> ans;
			ans.clear();
			int v = p1;
			for(int i=cnt-1;i>=1;i--)
			{
				int nextv = pre[i][v];
				int usedv = v - nextv;
				if(usedv == a[i][0])
				{
					for(int j=0;j<a[i][0];j++)
					{
						ans.push_back(b[i][0][j]);
					}
				}
				else
				{
					for(int j=0;j<a[i][1];j++)
					{
						ans.push_back(b[i][1][j]);
					}
				}
				v = nextv;
			}
			sort(ans.begin(),ans.end());
			for(int i=0;i<ans.size();i++)
			{
				printf("%d\n",ans[i]);
			}

			printf("end\n");

		}

	}
    return 0;
}