poj1417 带权并查集 + 背包 + 记录路径

True Liars

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 2713		Accepted: 868

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

 

题意: 有n个人，p1个好人p2个坏人。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。这里就可以用带权并查集处理。然后就是现在有k个集合了，每个集合中有坏人和好人，怎么组合才能满足要求。这时候就可以用背包来处理。dp[i][j]表示第i个集合后是否有j个人。dp[k][p1] = 1表示存在且唯一，dp[k][p1] = 0 表示不存在，dp[k][p1]>1表示多个。

然后关键在于怎么输出所有满足的人。这时候就可以在背包的时候记录路径，记录当前满足的状态是上面哪一个状态推过来的。

 

#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<string>
#include<time.h>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define INF 1000000001
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
const int MAXN = ;
int pa[MAXN],n,p1,p2,rel[MAXN],vis[MAXN];
vector<int>b[MAXN][];
int a[MAXN][],dp[MAXN][MAXN],pre[MAXN][MAXN];
void Init()
{
    for(int i = ; i <= p1 + p2; i++){
        pa[i] = i;
        b[i][].clear();
        b[i][].clear();
        a[i][] = ;
        a[i][] = ;
    }
    memset(rel,,sizeof(rel));
}
int find(int x)
{
    if(x != pa[x]){
        int fx = find(pa[x]);
        rel[x] = (rel[x] + rel[pa[x]]) % ;
        pa[x] = fx;
    }
    return pa[x];
}
int main()
{
    while(~scanf("%d%d%d",&n,&p1,&p2)){
        if(!n && !p1 && !p2)break;
        Init();
        int x,y,z;
        char s[];
        for(int i = ; i < n; i++){
            scanf("%d %d %s",&x,&y,s);
            if(s[] == 'y'){
                z = ;
            }
            else {
                z = ;
            }
            int fx = find(x);
            int fy = find(y);
            if(fx != fy){
                pa[fx] = fy;
                rel[fx] = ( - rel[x] + z + rel[y]) % ;
            }
        }
        memset(vis,,sizeof(vis));
        int cnt = ;
        for(int i = ; i <= p1 + p2; i++){
            if(!vis[i]){
                int tp = find(i);
                for(int j = i; j <= p1 + p2; j++){
                    int fp = find(j);
                    if(fp == tp){
                        vis[j] = ;
                        b[cnt][rel[j]].push_back(j);
                        a[cnt][rel[j]] ++;
                    }
                }
                //cout<<a[cnt][0]<<' '<<a[cnt][1]<<endl;
                cnt ++;
            }
        }
        memset(vis,,sizeof(vis));
        memset(dp,,sizeof(dp));
        dp[][] = ;
        for(int i = ; i < cnt; i++){
            for(int j = p1; j >= ; j--){
                if(j - a[i][] >=  && dp[i-][j - a[i][]]){
                    dp[i][j] += dp[i-][j-a[i][]];
                    pre[i][j] = j - a[i][];
                }
                if(j - a[i][] >=  && dp[i-][j - a[i][]]){
                    dp[i][j] += dp[i-][j-a[i][]];
                    pre[i][j] = j - a[i][];
                }
            }
        }
        if(dp[cnt-][p1] != ){
            printf("no\n");
        }
        else {
            vector<int>ans;
            int l = p1;
            for(int i = cnt - ; i >= ; i--){
                int tp = l - pre[i][l];
                if(tp == a[i][]){
                    for(int j = ; j < b[i][].size(); j++){
                        ans.push_back(b[i][][j]);
                    }
                }
                else {
                    for(int j = ; j < b[i][].size(); j++){
                        ans.push_back(b[i][][j]);
                    }
                }
                l = pre[i][l];
            }
            sort(ans.begin(),ans.end());
            for(int i = ; i < ans.size(); i++){
                printf("%d\n",ans[i]);
            }
            printf("end\n");
        }
    }
    return ;
}