题目链接：http://poj.org/problem?id=1417
Time Limit: 1000MS Memory Limit: 10000K
Description
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
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
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：
有一个人流落荒岛，岛上有好人阵营和坏人阵营，但是他没有办法从外表进行分别；
然后他决定问岛上的人X一个问题，问Y是不是好人；
好人阵营的人必须说真话，坏人阵营必须说谎话；
现在他只知道岛上好人阵营有p1个人，坏人阵营有p2个人，想要问n个问题，期望能确定每个人是否为好人……
若不能，则输出no；若能，则输出输出好人阵营的所有人的编号。
题解：
问X这个人Y是不是好人……
若得到的回答是NO：假设X是好人，则Y是坏人；假设X是坏人，则Y是好人。说明X和Y必然是敌对阵营的。
相反的，若得到的是YES答案，则X和Y是同一阵营的。
用并查集建树，每个节点有两个参数，par[x]和val[x]，par[x]表示x的父亲节点的编号，val[x]表示x与父亲节点是不是同一阵营；
val[x]为0表示该节点与其父亲节点是同阵营，val[x]为1表示该节点与其父亲节点是敌对阵营。
则用并查集对n个问题进行建树处理完之后，就将p1+p2个人分为k棵树（k个集合），每个集合上有a个人是一个阵营，b个是另一阵营，(但不确定哪个是好人阵营，哪个是坏人阵营)；
现在我们要做的就是构建一个方案：对k个集合，每个集合里选择a或b，认为它是这个集合内好人的人数，然后全部加起来正好等于p1；
若有且仅有这样的一个方案，那么就能精确判断那些人是好人阵营的，否则就只能输出no了。
AC代码：
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
const int maxn=++; int n,p1,p2; int par[maxn];
bool val[maxn];
void init(int l,int r)
{
for(int i=l;i<=r;i++) par[i]=i,val[i]=;
} int find(int x)
{
if(par[x]==x) return x;
else //下面三步非常关键，顺序不能颠倒，要理解val[x]跟随par[x]变动的意义
{
int tmp=find(par[x]);
val[x]=val[x]^val[par[x]];
return par[x]=tmp;
}
} struct Tree{
int a,b;
vector<int> anode;
vector<int> bnode;
}tree[maxn];
bool treeRootAppeared[maxn];
int treeID[maxn]; struct DP{
int val;
int pre;
}dp[maxn][maxn]; int main()
{
while(scanf("%d%d%d",&n,&p1,&p2) && n+p1+p2!=)
{
init(,p1+p2);
for(int i=,a,b;i<=n;i++)
{
char tmp[];
scanf("%d%d%s",&a,&b,tmp); int t1=find(a),t2=find(b);
if(t1!=t2)
{
par[t1]=t2;
if(tmp[]=='n') val[t1]=!(val[a]^val[b]); //a,b两者是敌对阵营
else val[t1]=val[a]^val[b]; //a,b两者是同阵营
}
} //对每个集合内的两个阵营人数进行统计
for(int i=;i<=p1+p2;i++) //初始化每个集合
{
tree[i].a=tree[i].b=;
tree[i].anode.clear();
tree[i].bnode.clear();
}
int cnt=;
memset(treeRootAppeared,,sizeof(treeRootAppeared));
for(int i=;i<=p1+p2;i++)
{
int t=find(i); if(!treeRootAppeared[t]) //这个集合的树根还未出现过
{
treeRootAppeared[t]=;
if(val[i]==)
{
tree[cnt].a++;
tree[cnt].anode.push_back(i);
}
else
{
tree[cnt].b++;
tree[cnt].bnode.push_back(i);
}
treeID[i]=treeID[t]=cnt;
cnt++;
}
else //这个集合的树根出现过了
{
if(val[i]==)
{
tree[treeID[t]].a++;
tree[treeID[t]].anode.push_back(i);
}
else
{
tree[treeID[t]].b++;
tree[treeID[t]].bnode.push_back(i);
}
treeID[i]=treeID[t];
}
} //动态规划求方案数
memset(dp,,sizeof(dp));
dp[][].val=;
for(int i=;i<cnt;i++)
{
for(int j=;j<=p1;j++)
{
if(j>=tree[i].a && dp[i-][j-tree[i].a].val>)
{
dp[i][j].val+=dp[i-][j-tree[i].a].val;
dp[i][j].pre=j-tree[i].a;
}
if(j>=tree[i].b && dp[i-][j-tree[i].b].val>)
{
dp[i][j].val+=dp[i-][j-tree[i].b].val;
dp[i][j].pre=j-tree[i].b;
}
}
} if(dp[cnt-][p1].val!=) printf("no\n");
else
{
vector<int> ans;
for(int i=cnt-,j=p1; i>=; j=dp[i][j].pre,i--)
{
int tmp=j-dp[i][j].pre; //tmp为第i个集合内好人的人数
if(tmp==tree[i].a)
for(int k=;k<tree[i].anode.size();k++) ans.push_back(tree[i].anode[k]);
else
for(int k=;k<tree[i].bnode.size();k++) ans.push_back(tree[i].bnode[k]);
} sort(ans.begin(),ans.end());
for(int i=;i<ans.size();i++) printf("%d\n",ans[i]);
printf("end\n");
}
}
}