POJ 1417 True Liars（并查集+DP）

True Liars

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 1556		Accepted: 457

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

Source

Japan 2002 Kanazawa

 

 

 

那么如果一个人说另一个人是好人，那么如果这个人是好人，说明 对方确实是好人，如果这个是坏人，说明这句话是假的，对方也是坏人。

如果一个人说另一个人是坏人，那么如果这个人是好人，说明对方是坏人，如果这个是坏人，说明 对方是好人。

也就是如果条件是yes说明这两个是相同集合的，否则是两个不同的集合。

用r[i]表示i结点与根结点的关系，0为相同集合，1为不同集合。这是一个经典的并查集问题。

这样处理之后，还需要判断是否唯一

我们通过并查集，可以将所有人分为若干个集合，其中对于每一个集合，又分为两个集合（好人和坏人，但是不知道哪些是好人，哪些是坏人，我们只有相对关系）

接下来就是从所有大集合中的两个小集合取一个，组成好人集合，判断是否唯一。

背包问题，dp[i][j]表示前i个大集合，好人为j个的方案有多少种，或者dp[i][j]表示当前好人i个，坏人j个的情况有多少种

如果dp[cnt][p1]！=1说明方案不唯一，或者无解。

 

输出方案就是加个pre数组，从后往前递推呢。

 

/*

POJ 1417

*/

#include <stdio.h>

#include <algorithm>

#include <iostream>

#include <string.h>

#include <vector>

#include <string>

using namespace std;



const int MAXN=610;

int F[MAXN];

int val[MAXN];

int find(int x)

{

    if(F[x]==-1)return x;

    int tmp=find(F[x]);

    val[x]+=val[F[x]];

    val[x]%=2;

    return F[x]=tmp;

}

int a[MAXN][2];//a[i][0],a[i][1]表示每个大集合分成两部分的个数

vector<int>b[MAXN][2];

bool used[MAXN];

int dp[MAXN][MAXN/2];

int pre[MAXN][MAXN/2];

int main()

{

    int n,p1,p2;

    while(scanf("%d%d%d",&n,&p1,&p2)==3)

    {

        if(n==0 && p1==0 && p2==0)break;

        memset(F,-1,sizeof(F));

        memset(val,0,sizeof(val));

        int u,v;

        char str[10];

        while(n--)

        {

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

            int tmp;

            if(str[0]=='y')//相同

              tmp=0;

            else tmp=1;//相反

            int t1=find(u),t2=find(v);

            if(t1!=t2)

            {

                F[t1]=t2;

                val[t1]=(val[v]-val[u]+tmp+2)%2;

            }

        }

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

        {

            b[i][0].clear();

            b[i][1].clear();

            a[i][0]=0;

            a[i][1]=0;

        }

        memset(used,false,sizeof(used));

        int cnt=1;

        for(int i=1;i<=p1+p2;i++)

          if(!used[i])

          {

              int tmp=find(i);

              for(int j=i;j<=p1+p2;j++)

              {

                  if(find(j)==tmp)

                  {

                      used[j]=true;

                      b[cnt][val[j]].push_back(j);

                      a[cnt][val[j]]++;

                  }

              }

              cnt++;

          }

        memset(dp,0,sizeof(dp));

        dp[0][0]=1;

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

        {

            for(int j=p1;j>=0;j--)

            {

                if(j-a[i][0]>=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]>=0 && 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 t=p1;

            //printf("%d\n",cnt);



            for(int i=cnt-1;i>=1;i--)

            {

                int tmp=t-pre[i][t];

                //printf("%d\n",i);

                //printf("%d  %d\n",t,tmp);

                if(tmp==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]);

                }

                t=pre[i][t];

            }



            sort(ans.begin(),ans.end());

            for(int i=0;i<ans.size();i++)

              printf("%d\n",ans[i]);

            printf("end\n");

        }



    }

    return 0;

}