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
此题的第一个难点就是在将给定的信息转化为并查集
对于每次输入yes代表相同阵营 否则则必为相反阵营
只需要将有关系的元素纳入同一个并查集
并统计出各个有关系的不同阵营的人数
然后使用01背包对将所有阵营的人数分配到两个派系中
最后只要统计所要凑成的方法数唯一即可
并不难的题自己用零散时间打了三天才写完
反映出之前所掌握的知识不够透彻
最近在写图论专题没回顾以前的知识也是一方面
期间wa的原因是把dp数组开成了1维
这样的做法是不对的
因为对于每一次有能力的选择 不选1那必定选2 所以之前的dp数组不存在不取操作
所以此题并不能将dp数组当成1维操作
附上我的ac代码
#include<cstdio>
#include<map>
#include<cstring>
#include<iostream>
#include<queue>
using namespace std;
int father[1055];
int rel[1055];//0 相同为零 不同为1
map<int,int> mp;
int bag[1055][2];
int dp[1055][1055];
int matr[1055][1055];
vector<int> ve[1005][2];
priority_queue <int,vector<int>,greater<int> > q;
int fin(int x)
{
if(father[x]==x) return x;
else
{
int tmp=father[x];
father[x]=fin(father[x]);
rel[x]=rel[x]^rel[tmp];
}
return father[x];
}
int main()
{
//freopen("in.txt","r",stdin);
int n,p1,p2;
while(~scanf("%d%d%d",&n,&p1,&p2))
{
// cas++;
// cout<<cas;
if(n==0&&p1==0&&p2==0) return 0;
int sum=p1+p2;
while(q.size()) q.pop();
memset(matr,0,sizeof(matr));
memset(dp,0,sizeof(dp));
memset(rel,0,sizeof(rel));
mp.clear();
memset(bag,0,sizeof(bag));
for(int i=1;i<=sum;i++)
{
for(int j=0;j<=1;j++)
ve[i][j].clear();
}
for(int i=1; i<=sum; i++)
{
father[i]=i;
}
int temp1,temp2;
char temp3[10];
for(int i=1; i<=n; i++)
{
scanf("%d%d%s",&temp1,&temp2,temp3);
int fa1=fin(temp1);
int fa2=fin(temp2);
if(temp1==temp2) continue;
father[fa2]=fa1;
if(temp3[0]=='y')
{ if(rel[temp1]!=rel[temp2]) rel[fa2]=1; }
else
{ if(rel[temp1]==rel[temp2]) rel[fa2]=1; }
}
int cnt=0;
for(int i=1; i<=sum; i++)
{
int tmp=fin(i);
//cout<<"/"<<tmp<<endl;
if(mp[tmp]==0) mp[tmp]=++cnt;
bag[mp[tmp]][rel[i]]++;
ve[mp[tmp]][rel[i]].push_back(i);
}
// for(int i=1;i<=cnt;i++)
// {
// cout<<"*";
// cout<<bag[i][1]<<" "<<bag[i][0]<<endl;
// }
// //背包部分
dp[0][0]=1;
matr[0][0]=1;
for(int i=1; i<=cnt; i++)
{
for(int j=p1; j>=0; j--)
{
if(j>=bag[i][0])
dp[i][j]+=dp[i-1][j-bag[i][0]];//每个集合对应两种状态
if(j>=bag[i][1])
dp[i][j]+=dp[i-1][j-bag[i][1]];
}
}
int pos=p1;
// printf("%d
",dp[p1]);
if(dp[cnt][p1]==1)
{
for(int i=cnt; i>=1; i--)
{
// if(matr[i][tmp-bag[i][0]]==1&&bag[i][0]!=0)
// cout<<"1"<<i<<endl;
// else if(matr[i][tmp-bag[i][1]]==1&&bag[i][1]!=0)
// cout<<"2"<<i<<endl;
if(dp[i-1][pos-bag[i][0]]==1)
{
// cout<<i<<"0"<<endl;
pos=pos-bag[i][0];
for(int j=0;j<ve[i][0].size();j++)
{
q.push(ve[i][0][j]);
}
}
else if(dp[i-1][pos-bag[i][1]]==1)
{
//cout<<i<<"1"<<endl;
pos=pos-bag[i][1];
for(int j=0;j<ve[i][1].size();j++)
{
q.push(ve[i][1][j]);
}
}
}
while(q.size())
{
printf("%d
",q.top());
q.pop();
}
printf("end
");
}
else
printf("no
");
}
}