True Liars
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 3188   Accepted: 1026

Description

After having drifted about in a small boat for a couple of days, Akira Crusoe Maeda was finally cast ashore on a foggy island. Though he was exhausted and despaired, he was still fortunate to remember a legend of the foggy island, which he had heard from patriarchs in his childhood. This must be the island in the legend. In the legend, two tribes have inhabited the island, one is divine and the other is devilish, once members of the divine tribe bless you, your future is bright and promising, and your soul will eventually go to Heaven, in contrast, once members of the devilish tribe curse you, your future is bleak and hopeless, and your soul will eventually fall down to Hell. 

In order to prevent the worst-case scenario, Akira should distinguish the devilish from the divine. But how? They looked exactly alike and he could not distinguish one from the other solely by their appearances. He still had his last hope, however. The members of the divine tribe are truth-tellers, that is, they always tell the truth and those of the devilish tribe are liars, that is, they always tell a lie. 

He asked some of them whether or not some are divine. They knew one another very much and always responded to him "faithfully" according to their individual natures (i.e., they always tell the truth or always a lie). He did not dare to ask any other forms of questions, since the legend says that a devilish member would curse a person forever when he did not like the question. He had another piece of useful informationf the legend tells the populations of both tribes. These numbers in the legend are trustworthy since everyone living on this island is immortal and none have ever been born at least these millennia. 

You are a good computer programmer and so requested to help Akira by writing a program that classifies the inhabitants according to their answers to his inquiries. 

Input

The input consists of multiple data sets, each in the following format : 

n p1 p2 
xl yl a1 
x2 y2 a2 
... 
xi yi ai 
... 
xn yn an 

The first line has three non-negative integers n, p1, and p2. n is the number of questions Akira asked. pl and p2 are the populations of the divine and devilish tribes, respectively, in the legend. Each of the following n lines has two integers xi, yi and one word ai. xi and yi are the identification numbers of inhabitants, each of which is between 1 and p1 + p2, inclusive. ai is either yes, if the inhabitant xi said that the inhabitant yi was a member of the divine tribe, or no, otherwise. Note that xi and yi can be the same number since "are you a member of the divine tribe?" is a valid question. Note also that two lines may have the same x's and y's since Akira was very upset and might have asked the same question to the same one more than once. 

You may assume that n is less than 1000 and that p1 and p2 are less than 300. A line with three zeros, i.e., 0 0 0, represents the end of the input. You can assume that each data set is consistent and no contradictory answers are included. 

Output

For each data set, if it includes sufficient information to classify all the inhabitants, print the identification numbers of all the divine ones in ascending order, one in a line. In addition, following the output numbers, print end in a line. Otherwise, i.e., if a given data set does not include sufficient information to identify all the divine members, print no in a line.

Sample Input

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

Sample Output

no
no
1
2
end
3
4
5
6
end


题目大意：给你p1个好人和p2个坏人，编号为1-p1+p2，然后给你n中操作

                  x1 x2 no：x1说x2不是好人

                  x1 x2 yes：x1说x2是好人

                  在这里好人说的总是对的，坏人说的总是坏的，然后问你最后能不能唯一确定哪些是好人，并输出不能就输出no

题目思路：开始看到这题很好想到的就是用并查集，如果是yes的话说明他们同类，no的话说明不是同类，但这样只能分出几个大集                  

                  合来，每个集合又分成两个小集合表示两种类型的个数，而我们要求的是在所有大集合中选出一个小集合然后加起来看能                   

                 不能组 合成p1，并且要唯一，这里我们可以想到用背包来做，dp[i][j]表示前i个大集合好人为j个的方案数，第i种状态只能

                  是有第i-1种状态而来，我们用w0[i],w1[i]表示第i个集合两个小集合的个数 ，

                  所以dp[i][j]可以由dp[i-1][j-w0[]i]和dp[i-1][j-w1[i]]得来，这样我们只需判断dp[cnt][p1]是否等于1，

                  这题还有麻烦的就是输出好人的编号，这里我们可以利用边的权值，w0存的全是权值为0的w1存的全是权值为1的，然后                  

                   最后只需判断下第i个集合是由w0还是w1组合而来，这里可以利用dp[i-1][p1-w0[i]]的值来判断，如果等于1则表示是由w0                     

                   组合否则就是w1，这个不难理，dp[i-1][p1-w0[i]]==1可以得到dp[i][p1]==1,应为这里已经表示答案唯一，所以只有一种情                    

                     况！

AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

int pre[1000],rk[1000],p[1000],vis[1000];
int w0[1000],w1[1000];
int dp[700][700];
int n,p1,p2;

int get(int x){
    if(x!=pre[x]){
        int tmp = pre[x];
        pre[x] = get(pre[x]);
        rk[x] = rk[x]^rk[tmp];
    }
    return pre[x];
}

void unio(int x,int y,int k){
    int xx =get(x);
    int yy = get(y);
    if(xx!=yy){
        pre[xx] = yy;
        rk[xx]=rk[x]^rk[y]^k;
    }
}

int main()
{
    while(scanf("%d%d%d",&n,&p1,&p2)&&(n+p1+p2)){
        for(int i=1;i<=p1+p2;i++)
            pre[i]=i,rk[i]=0,vis[i]=0,w0[i]=0,w1[i]=0;
        memset(dp,0,sizeof(dp));
        for(int i=0;i<n;i++){
            int a,b;char s[10];
            scanf("%d%d%s",&a,&b,s);
            if(s[0]=='y')unio(a,b,0);
            else unio(a,b,1);
        }
        int cnt = 1;
        for(int i=1;i<=p1+p2;i++){
            if(!vis[i]){
                int f = get(i);
                for(int j=i;j<=p1+p2;j++){
                    if(get(j)==f&&!vis[j]){
                        vis[j]=1;
                        if(rk[j]==0)w0[cnt]++;
                        else w1[cnt]++;
                    }
                }
                p[cnt]=f;
                cnt++;
            }
        }
        dp[0][0]=1;
        for(int i=1;i<cnt;i++){
            int Min=min(w0[i],w1[i]);
            for(int j=p1;j>=Min;j--){
                if(dp[i-1][j-w0[i]]){
                    dp[i][j]+=dp[i-1][j-w0[i]];
                }
                if(dp[i-1][j-w1[i]]){
                    dp[i][j]+=dp[i-1][j-w1[i]];
                }
            }
        }
        if(dp[cnt-1][p1]!=1){
            printf("no\n");
            continue;
        }
        int ans[1000];
        int num = 0;
        int Mp = p1;
        for(int i=cnt-1;i>=1;i--){
            if(dp[i-1][Mp-w0[i]]==1){
                for(int j=1;j<=p1+p2;j++){
                    if(get(j)==p[i]&&rk[j]==0)ans[num++]=j;
                }
                Mp-=w0[i];
            }
            else {
                 for(int j=1;j<=p1+p2;j++){
                    if(get(j)==p[i]&&rk[j]==1)ans[num++]=j;
                }
                Mp-=w1[i];
            }
        }
        sort(ans,ans+num);
        for(int i=0;i<num;i++)
            printf("%d\n",ans[i]);
        puts("end");
    }
    return 0;
}