True Liars
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 2221 | Accepted: 685 |
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
分析：题意：在一个孤岛上有两个部落，一个部落都是好人，一个部落都是坏人，好人都说真话，坏人
都说假话。有一群人来自两个部落，通过给定的数据确定那些是好人，且以升序输出。
这道题有一点需要注意：好人说真话，如果一个人是好人，他说另一个人是坏人这这个人一定是坏人，否则
这个人一定是好人；相反，如果一个人是坏人，他说另一个人是坏人则这个人一定是好人，否则这个人一定
是坏人。另外要记住在这里面好人坏人是相对的，也就是说我们只能判断谁和谁同等人，即yes带表同等集合
no表示相反集合。
这道题我们要最终把这一群人分成若干集合（一颗树就是一个集合，一群人是一个森林），然后每一个集合分成
朋友，敌人两个子集a和b（大家不要误解朋友集合和敌人集合的意思，它们是通过并查集确定的两个集合，在这两个集合中
必有一个集合为好人，一个集合为坏人，只不过我们现在还不确定是好谁坏）。这个用并查集可以不难做到。难点是一下：
因为题目已经告诉好人坏人的个数，因此我们从这若干个集合中每个集合挑出一个子集，是这些子集之和等于好人
数目p1，判断这样的情况有多少种，如果答案唯一则可以确定那些事好人，那些是坏人了，因此需要用
到动态规划中的背包一类。
设数组dp【maxn】【maxn】，dp【i】【j】代表到第i个集合好人数为j的情况数; 状态转移方程为：
dp【i】【j】=dp【i-1】【j-a】+dp【i-1】【j-a】；
如果dp【cnt】【p1】==1说明情况唯一（cnt为集合数），则可一找出好人个数；
由于小生是dp学的不好所以猥琐看了某君dp代码，其中有一句：点dp【i】【j】>1时我都是dp【i】【j】=2；
以为算到最后如果dp【cnt】【p1】>1就无解，所以dp【i】【j】>1就行。
唉，必须好好学啊，加油。
献上代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
const int maxn=600+10;
int fa[maxn],level[maxn],id[maxn];
int dp[maxn][maxn],ans[maxn][maxn],map[maxn][2],num[maxn][2];
void Init(int snum)
{
for(int i=0;i<=snum;i++)
{
fa[i]=i;
}
memset(dp,0,sizeof(dp));
memset(level,0,sizeof(level));
memset(ans,0,sizeof(ans));
memset(map,0,sizeof(map));
memset(num,0,sizeof(num));
memset(id,0,sizeof(id));
}
int find(int x)
{
if(x!=fa[x])
{
int tmp=fa[x];
fa[x]=find(fa[x]);
level[x]=level[x]^level[tmp];
}
return fa[x];
}
int main()
{
int n,p1,p2,x,y,fx,fy,cnt;
char c[10];
while(~scanf("%d%d%d",&n,&p1,&p2)&&(n+p1+p2))
{
Init(p1+p2);
for(int i=0;i<n;i++)
{
scanf("%d%d%s",&x,&y,c);
fx=find(x);
fy=find(y);
if(fx==fy)
continue;
fa[fy]=fx;
if(c[0]==‘y‘)
{
level[fy]=level[x]^level[y]^0;
}
else
{
level[fy]=level[x]^level[y]^1;
}
}
cnt=0;
for(int i=1;i<=p1+p2;i++)
if(find(i)==i)id[i]=++cnt;
for(int i=1;i<=p1+p2;i++)
{
map[id[find(i)]][level[i]]++;
}
dp[0][0]=1;
for(int i=1;i<=cnt;i++)
for(int j=0;j<=p1+p2;j++)
{
if(j>=map[i][0]&&dp[i-1][j-map[i][0]]>0)
{
dp[i][j]+=dp[i-1][j-map[i][0]];
ans[i][j]=map[i][0];
}
if(j>=map[i][1]&&dp[i-1][j-map[i][1]]>0)
{
dp[i][j]+=dp[i-1][j-map[i][1]];
ans[i][j]=map[i][1];
}
if(dp[i][j]>1)
dp[i][j]=2;
}
if(dp[cnt][p1]!=1)
{
printf("no\n");
continue;
}
for(int i=cnt,j=p1;i>=1;i--)
{
if(map[i][0]==ans[i][j])num[i][0]=1;
else num[i][1]=1;
j-=ans[i][j];
}
for(int i=1;i<=p1+p2;i++)
{
if(num[id[find(i)]][level[i]])
printf("%d\n",i);
}
printf("end\n");
}
return 0;
}