【POJ1417】【带标记并查集+DP】True Liars

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

【分析】

比较简单的题目，前面先用类似食物链的方法得到一些并查集集合，然后将每个并查集里面的与根在同一个集合中和不再同一个集合中的统计出来。

然后就就变成在每个并查集中从两个数中选一个数成为特定的值，很简单的DP。

最后输出方案简单标记一下，逆推即可。

  1 #include <iostream>

  2 #include <cstdio>

  3 #include <algorithm>

  4 #include <cstring>

  5 #include <vector>

  6 #include <utility>

  7 #include <iomanip>

  8 #include <string>

  9 #include <cmath>

 10 #include <queue>

 11 #include <map>

 12 #define LOCAL

 13 const int MAXN = 1000 + 10; 

 14 const int MAX = 30000 + 10; 

 15 using namespace std;

 16 struct DATA{

 17        int a, b;

 18 }data[MAXN];

 19 int n, p1, p2, q;

 20 int parent[MAXN] ,val[MAXN];

 21 int num[MAXN];//num[i]代表如果i是并查集中的根，那么他在AB表中的位置就num[i] 

 22 int f[MAXN][MAXN], cnt; 

 23 int pre[MAXN][MAXN];//DP中记录状态转移过来的位置 

 24 int Ans[MAXN];//表示答案中被选为好人或者不是好人 

 25 

 26 int find(int x){

 27     int f = x, tmp = 0;

 28     while (x != parent[x]){

 29           tmp += val[x];

 30           x = parent[x];

 31     }

 32     parent[f] = x;

 33     val[f] = tmp % 2;

 34     return x;

 35 }

 36 

 37 void init(){

 38      n = p1 + p2;//即总的人数 

 39      //val为0代表相同集合，1代表不同的集合 

 40      for (int i = 1; i <= n; i++){

 41          val[i] = 0;

 42          parent[i] = i; 

 43      }

 44      for (int i = 1; i <= q; i++){

 45          int x, y;

 46          char str[5];

 47          scanf("%d%d", &x, &y);

 48          scanf("%s", str);

 49          if (find(x) == find(y)) continue;//在同一个集合之内就不用考虑了 

 50          else{

 51               if (str[0] == 'y'){ //一定是一种集合之内 

 52                  int xx = find(x), yy = find(y);

 53                  parent[xx] = y;

 54                  val[xx] = (-val[x] + 2) % 2;

 55               }else{

 56                  int xx = find(x), yy = find(y);

 57                  parent[xx] = y;

 58                  val[xx] = (-val[x] + 3) % 2; 

 59               }

 60          }

 61      }

 62 }

 63 //计算出各个并查集的同类和不同类 

 64 void prepare(){

 65      memset(data, 0, sizeof(data));

 66      cnt = 0;

 67      for (int i = 1; i <= n; i++){

 68          if (parent[i] != i) continue;

 69          num[i] = ++cnt;

 70      }

 71      for (int i = 1; i <= n; i++){

 72          int xx = find(i);

 73          if (val[i] == 0) data[num[xx]].a++;//和它同一类

 74          else data[num[xx]].b++;//不同类 

 75      }

 76 }

 77 void dp(){

 78      memset(f, 0, sizeof(f));

 79      //f[i][j]表示到了第i个好人有j个的时候的方案数量,注意只要保存3个量就可以了 

 80      f[0][0] = 1;

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

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

 83          if (j - data[1].a >= 0) f[i][j] += f[i - 1][j - data[i].a];

 84          if (j - data[1].b >= 0) f[i][j] += f[i - 1][j - data[i].b];

 85          if (f[i][j] >= 2)  f[i][j] = 2;//不要做太大了 

 86          if (f[i][j] == 1){//有解

 87             if (j - data[i].a >= 0 && f[i - 1][j - data[i].a] == 1) pre[i][j] = 0;//a类选为好人 

 88             if (j - data[i].b >= 0 && f[i - 1][j - data[i].b] == 1) pre[i][j] = 1;//b类选为好人 

 89          }

 90      } 

 91      if (f[cnt][p1] != 1){printf("no\n");return;}

 92      int last = p1;

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

 94          Ans[i] = pre[i][last];

 95          if (Ans[i] == 0) last -= data[i].a;

 96          else last -= data[i].b;

 97      }

 98      for (int i = 1; i <= n; i++){

 99          int xx = find(i);

100          if (Ans[num[xx]] == 1 && val[i] == 1) printf("%d\n", i);            

101          if (Ans[num[xx]] == 0 && val[i] == 0) printf("%d\n", i);

102      }

103      printf("end\n");

104 } 

105  

106 int main(){

107     int T;

108     #ifdef LOCAL

109     freopen("data.txt",  "r",  stdin);

110     freopen("out.txt",  "w",  stdout); 

111     #endif

112     while (scanf("%d%d%d", &q, &p1, &p2)){

113           if (q == 0 && p1 == 0 && p2 == 0) break;

114           init();

115           prepare();

116           dp();

117           //printf("%d", data[1].b); 

118     }

119     return 0;

120 }

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