Problem link: 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
Title:
There is a man who lives on a deserted island. There are good people and bad people on the island, but he has no way to distinguish from appearance;
Then he decided to ask the islander X a question and asked if Y was a good person;
Those in the good camp must tell the truth, and those in the bad camp must lie;
Now he only knows that there are p1 people in the good camp and p2 people in the bad camp on the island. He wants to ask n questions, hoping to determine whether everyone is a good person...
If not, output no; if yes, output the numbers of all the people in the good camp.
answer:
Ask X if this person Y is a good person...
If the answer is NO: if X is a good person, then Y is a bad person; if X is a bad person, then Y is a good person. Explain that X and Y must be in hostile camps.
Conversely, if the answer is YES, then X and Y are in the same camp.
Use union search to build a tree, each node has two parameters, par[x] and val[x], par[x] represents the number of the parent node of x, and val[x] represents whether x and the parent node are in the same camp;
If val[x] is 0, it means that the node is in the same camp as its parent node, and if val[x] is 1, it means that the node and its parent node are in the opposite camp.
Then use the union search set to build a tree and deal with the n problems, divide p1+p2 people into k trees (k sets), each set has a person who is a camp, b is another camp, (but not sure which is the good camp and which is the bad camp);
Now what we have to do is to construct a scheme: for k sets, choose a or b in each set, consider it to be the number of good people in this set, and then add them all up exactly equal to p1;
If there is one and only such a plan, then it can be accurately judged that those people are in the good camp, otherwise it can only output no.
AC code:
#include<cstdio> #include<cstring> #include<algorithm> #include<vector> using namespace std; const int maxn=300+300+10; 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]=0; } int find(int x) { if (par[x]==x) return x; else // The following three steps are very critical, the order cannot be reversed, to understand the meaning of val[x] following par[x] changes { 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]; intmain () { while(scanf("%d%d%d",&n,&p1,&p2) && n+p1+p2!=0) { init(1,p1+p2); for(int i=1,a,b;i<=n;i++) { char tmp[5]; scanf("%d%d%s",&a,&b,tmp); int t1=find(a),t2=find(b); if(t1!=t2) { par[t1] = t2; if (tmp[ 0 ]== ' n ' ) val[t1]=!(val[a]^val[b]); // both a and b are hostile factions else val[ t1]=val[a]^val[b]; // a and b are in the same camp } } // Count the number of two camps in each set for ( int i= 1 ;i<=p1+p2;i++) // initialize each set { tree[i].a=tree[i].b=0; tree[i].anode.clear(); tree[i].bnode.clear(); } int cnt=1; memset(treeRootAppeared,0,sizeof(treeRootAppeared)); for(int i=1;i<=p1+p2;i++) { int t=find(i); if (!treeRootAppeared[t]) // The tree root of this set has not yet appeared { treeRootAppeared[t]=1; if(val[i]==0) { 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 // The root of this set has appeared { if (val[i]== 0 ) { 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]; } } // Dynamic programming to find the number of solutions memset(dp, 0 , sizeof (dp)); dp[0][0].val=1; for(int i=1;i<cnt;i++) { for(int j=0;j<=p1;j++) { if(j>=tree[i].a && dp[i-1][j-tree[i].a].val>0) { dp[i][j].val+=dp[i-1][j-tree[i].a].val; dp[i][j].pre=j-tree[i].a; } if(j>=tree[i].b && dp[i-1][j-tree[i].b].val>0) { dp[i][j].val+=dp[i-1][j-tree[i].b].val; dp[i][j].pre=j-tree[i].b; } } } if(dp[cnt-1][p1].val!=1) printf("no\n"); else { vector<int> ans; for(int i=cnt-1,j=p1; i>=1; 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=0;k<tree[i].anode.size();k++) ans.push_back(tree[i].anode[k]); else for(int k=0;k<tree[i].bnode.size();k++) ans.push_back(tree[i].bnode[k]); } sort(ans.begin(),ans.end()); for(int i=0;i<ans.size();i++) printf("%d\n",ans[i]); printf("end\n"); } } }