POJ 1417 并查集 dp

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

用并查集来处理数据，然后就变成了背包

带权并查集，权重为0或1
0代表和父节点同类，1代表和父节点不同类，
这里的同类关系并不能判断出是诚实的人还是说谎的人，需要在后面dp判断。

首先，将yes视作同类关系，no视作异类关系。
然后如果x，y同根的话在判断一下是否矛盾，不过貌似数据里没有。
处理出来一共有多少个集合，并且把集合的根存储起来，还要处理出根的同类，根的异类的数目。
接着就是dp了。
因为每一个集合都必须要用到，所以dp[i][0]不能初始化为1。
所以使用第一个集合来初始化dp的起点，另外初始化的时候要注意是 += ，
因为有可能第一个集合的同类与异类的数目一样，当然这样的话是没法唯一的表示出t和l的(也有可能表示不出来）
dp完成之后判断一下dp[k-1][t]（dp[k-1][l]也是一样的）是否等于1，等于的话就能唯一的表示，不能的话no（不论是等于0还是大于一）
接着就是输出路径了。

#include<cstdio>
#include<cstring>

int dp[][];
struct s{
    int father, relation;
    int same, other;
    int True;
}p[];

int find_(int x){
    if(x == p[x].father)
        return x;
    int px = p[x].father;
    p[x].father = find_(px);
    p[x].relation = (p[x].relation + p[px].relation)%;
    return p[x].father;
}

int main(){
    int m, t, l;
    int x, y;
    char str[];
    while(scanf("%d%d%d",&m,&t,&l),m|t|l){
        int n = t+l, d, fx, fy, ok = , k = ;
        int f[];
        for(int i=;i<=n;i++){
           p[i] = s{i,,,,};
           // 自己和自己的关系是同类 0
           // 自己和自己是同类，所以same = 1
        }
        for(int i=;i<m;i++){
            scanf("%d%d%s",&x,&y,str);
            if(ok)continue;
            if(str[] == 'y') // 同类
                d = ;
            else d = ;
            fx = find_(x);
            fy = find_(y);

            if(fx == fy && (p[x].relation + p[y].relation)% != d)
                ok = ;
            else {
                p[fx].father = fy;
                p[fx].relation = (p[x].relation+p[y].relation+d)%;
            }
        }
        if(!ok){
            for(int i=;i<=n;i++){
                x = find_(i);
                if(x == i)
                    f[k++] = i;
                else {
                    p[x].other += p[i].relation;
                    p[x].same +=  - p[i].relation;
                }
            }

            memset(dp,,sizeof(dp));

            dp[][ p[f[]].same ] += ;
            dp[][ p[f[]].other ] += ;
            for(int i=;i<k;i++){
                x = f[i];
                for(int j=;j<=n;j++){
                    if(dp[i-][j]){
                        dp[i][ p[x].same + j ] += dp[i-][j];
                        dp[i][ p[x].other + j ] += dp[i-][j];
                    }
                }
            }
        }
        if(dp[k-][t] !=  || ok)
            printf("no\n");
        else {
            for(int i=k-;i>;i--){
                x = f[i];
                y = p[x].same;
                if(i !=  && dp[i-][t-y] !=  || (i ==  && t == y)){
                    p[x].True = ;
                    t -= y;
                }else t -= p[x].other;
            }
            for(int i=;i<=n;i++){
                x = p[i].father;
                if(p[x].True && !p[i].relation || p[x].True == && p[i].relation)
                    printf("%d\n",i);
            }
            printf("end\n");
        }
    }
    return ;
}