True Liars （并查集压缩路径 + DP）

                                     True Liars 

题目的连接

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

题意：有两个部落，一个天使部落，有p1 ren，一个魔鬼部落，有p2人，所有人都有编号，从1 到 p1+p2，魔鬼部落说的都是谎话 ，天使部落说的都是真话，当为yes时，如果a是天使，他的话为真，b一定是天使，若a为魔鬼时，那么他的话为假，所以b一定也是魔鬼，当为no时，如果a为天使，他的话为真，所以，b一定是魔鬼，若a为魔鬼，a的话为假，所以b为天使，所以当为yes时，a 与 b 是同一个部落的 ， 为 no 时 , a 与 b 一定不是同一个部落 。

分析：事实上这个题目前面的并查集部分只是一个普通的种类并查集，这个记录路径的dp才是本题解的精妙部分；我们用vis[i]来表示与父结点的关系，0表示两者同类，1表示两者不同类。前面是不同的带权并查集，，你可以得到很多集合，每个集合都有两个子集合，通过查找dp来查找，每个大集合的一个小集合的个数是p1。w0[] w1[]存每个大集合里面连个小集合的个数。

通过dp[ i ][ j ] = sum(dp[ i - 1][ j - w1[ i ]] + dp[i - 1][ j  -  w0[ j ]] ) ;

如果dp[ cnt ][ p1 ] 不等于 1, 那么输出 no

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<cmath>

using namespace std;

const int Maxn = 1000;

int f[Maxn] , vis[Maxn];
string s;
int p[Maxn] , w1[Maxn] , w0[Maxn];
bool used[Maxn];

int dp[Maxn][Maxn];

int Find(int x){
    if(x != f[x]){
        int k = f[x];
        f[x] = Find(f[x]);
        vis[x] = (vis[x] + vis[k]) % 2;
    }
    return f[x];
}

int main(){
    int n , p1, p2 , a , b;
    while(scanf("%d%d%d",&n , & p1 , & p2) && n + p1 + p2 != 0){
        memset(dp,0,sizeof(dp));

        int  t = 0;
        for(int i = 0;i <= p1+p2;i++) f[i] = i , vis[i] = 0 , w1[i] = 0 ,w0[i] = 0, used[i] = false;       // 初始化

        while(n--){
            cin >> a >> b >> s;
            int ta = Find(a);
            int tb = Find(b);
            if(ta != tb){
                f[ta] = tb;
                if(s[0] == 'y') vis[ta] =  (vis[a] + vis[b]) % 2;
                else vis[ta] = (vis[a] + vis[b] + 1) % 2;
            }
        }
        // 求出每个集合包括两个小集合的个数
        int cnt = 1;
        for(int i = 1;i <= p1 + p2;i++){
            if(!used[i]){
                int t = Find(i);
                for(int j = i;j <= p1 + p2;j++){
                    if(Find(j) == t && !used[j]){
                        used[j] = true;
                        if(!vis[j]) w0[cnt]++;
                        else w1[cnt]++;
                    }
                }
                p[cnt] = t; // 将集合的根结点存起来
                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 flag= p1;
        for(int i = cnt - 1;i >= 1;i--){
            if(dp[i-1][flag - w0[i]] == 1){
                for(int j = 1;j <= p1 + p2;j++)
                  if(Find(j) == p[i] && vis[j] == 0) ans[num++] = j;
                flag -= w0[i];
            }
            else if(dp[i-1][flag - w1[i]] == 1){
                for(int j = 1;j <= p1 + p2;j++)
                  if(Find(j) == p[i] && vis[j] == 1) ans[num++] = j;
                flag -= w1[i];
            }
        }
        sort(ans ,ans + num);
        for(int i = 0;i < num;i++) printf("%d\n",ans[i]);
        cout << "end\n";
    }
    return 0;
}