POJ1417 True Liars —— 并查集 + DP

题目链接：http://poj.org/problem?id=1417

True Liars

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 3851		Accepted: 1261

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

 

 

 

题解：

1.题目分析：此题的弱化版 POJ2492 ， 做了就能明白怎么用并查集处理此题每个人之间的关系了。

2.对于每个集合，有属于0子集的， 有属于1子集的。我们要做的就是，判断是否仅存在一种情况使得在每个集合中选出一个子集，他们的元素个数为p1。怎么判断呢？动态规划的思想。

3.详情请看代码。

 

 

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
#define ms(a,b) memset((a),(b),sizeof((a)))
using namespace std;
typedef long long LL;
const double EPS = 1e-8;
const int INF = 2e9;
const LL LNF = 2e18;
const int MAXN = 1e3+10;

int n, m, p1, p2;
int fa[MAXN], r[MAXN];
int cnt, Set[MAXN], Map[MAXN], sum[MAXN][2];
int dp[MAXN][330], used[MAXN][2], chosen[MAXN][330];

int find(int x)
{
    if(fa[x]==x) return x;
    int pre = find(fa[x]);
    r[x] = (r[x]+r[fa[x]])%2;
    return fa[x] = pre;
}

void Union(int u, int v, int w)
{
    int fu = find(u);
    int fv = find(v);
    if(fu!=fv)
    {
        fa[fu] = fv;
        r[fu] = (-r[u]+w+r[v]+2)%2;
    }
}

int main()
{
    while(scanf("%d%d%d", &m, &p1, &p2) && (m||p1||p2))
    {
        n = p1 + p2;
        for(int i = 1; i<=n; i++)   //此题注意要把fa[i]初始化为i，否则的话根节点找不到fa
            fa[i] = i, r[i] = 0;
        for(int i = 1; i<=m; i++)
        {
            int u, v; char s[5];
            scanf("%d%d%s", &u, &v, s);
            Union(u, v, (int)(s[0]=='n'));
        }

        cnt = 0;
        for(int i = 1; i<=n; i++)   //统计集合个数。Set[]用于存储有哪些集合， Map[]用于离散化。Set和Map互逆
            if(fa[i]==i)
                Set[++cnt] = i, Map[i] = cnt;

        memset(sum, 0, sizeof(sum));
        for(int i = 1; i<=n; i++)   //统计每个集合的0、1个数
            sum[Map[find(i)]][r[i]]++;

        memset(dp, 0, sizeof(dp));
        dp[0][0] = 1;
        for(int i = 1; i<=cnt; i++)
        for(int j = 0; j<=p1; j++)
        {
            if(j>=sum[i][0] && dp[i-1][j-sum[i][0]])    //注意要判断dp[i-1][j-sum[i][0]]是否不等于0
            {
                dp[i][j] += dp[i-1][j-sum[i][0]];
                chosen[i][j] = 0;  //此集合选择了0
            }
            if(j>=sum[i][1] && dp[i-1][j-sum[i][1]])
            {
                dp[i][j] += dp[i-1][j-sum[i][1]];
                chosen[i][j] = 1;   //此集合选择了1
            }
        }

        if(dp[cnt][p1]!=1)
        {
            printf("no
");
            continue;
        }

        int num = p1;
        memset(used, 0, sizeof(used));
        for(int i = cnt; i>=1; i--)    //从结果倒推回去，标记每个集合实际用了0还是1
        {
            used[i][chosen[i][num]] = 1;
            num -= sum[i][chosen[i][num]];
        }

        vector<int>a;
        a.clear();
        for(int i = 1; i<=n; i++)
            if(used[Map[fa[i]]][r[i]])
                a.push_back(i);

        sort(a.begin(), a.end());
        for(int i = 0; i<a.size(); i++)
            printf("%d
", a[i]);
        printf("end
");
    }
}