Werewolf（狼人杀） is a game in which the players are partitioned into two parties: the werewolves and the human beings. Suppose that in a game,
- player #1 said: "Player #2 is a werewolf.";
- player #2 said: "Player #3 is a human.";
- player #3 said: "Player #4 is a werewolf.";
- player #4 said: "Player #5 is a human."; and
- player #5 said: "Player #4 is a human.".
Given that there were 2 werewolves among them, at least one but not all the werewolves were lying, and there were exactly 2 liars. Can you point out the werewolves?
Now you are asked to solve a harder vertion of this problem: given that there were N players, with M werewolves among them, at least one but not all the werewolves were lying, and there were exactly L liars. You are supposed to point out the werewolves.
Input Specification:
Each input file contains one test case. For each case, the first line gives three positive integer N (5 ≤ N ≤ 100), M and L (2 ≤ M，L < N). Then N lines follow and the i-th line gives the statement of the i-th player (1 ≤ i ≤ N), which is represented by the index of the player with a positive sign for a human and a negative sign for a werewolf.
Output Specification:
If a solution exists, print in a line in descending order the indices of the M werewolves. The numbers must be separated by exactly one space with no extra spaces at the beginning or the end of the line. If there are more than one solution, you must output the largest solution sequence -- that is, for two sequences A = { a[1], ..., a[M] } and B = { b[1], ..., b[M] }, if there exists 0 ≤ k < M such that a[i] = b[i] (i ≤ k) and a[k+1]>b[k+1], then A is said to be larger than B. In case there is no solution, simply print No Solution.
Sample Input 1:
5 2 2
-2
+3
-4
+5
+4
Sample Output 1:
4 1
Sample Input 2:
6 2 3
-2
+3
-4
+5
+4
-3
Sample Output 2:
6 4
Sample Input 3:
6 2 5
-2
+3
-4
+5
+4
+6
Sample Output 3:
No Solution1 #include<bits/stdc++.h> 2 using namespace std; 3 vector<int> v,vr,vl,vi; 4 int n,m,k; 5 void dfs(int n,int ln,int wn,int un) 6 { 7 if(ln>k||wn>m||wn+un<m) 8 return; 9 else if(n<k-ln) 10 return; 11 else if(n<1) 12 { 13 if(ln==k&&wn<=m&&wn+un>=m) 14 { 15 int first=0; 16 bool flag=false; 17 int fl=0; 18 for(int i=v.size()-1;i>0&&!flag;--i) 19 { 20 if(vl[i]==-1&&vi[i]==-1) 21 flag=true; 22 } 23 if(m-wn) 24 for(int i=v.size()-1;i>0&&!flag;--i) 25 { 26 if(vl[i]==-1&&vi[i]==0) 27 { 28 fl=i; 29 vi[i]=-1; 30 wn++; 31 un--; 32 flag=true; 33 } 34 } 35 if(flag) 36 { 37 vector<int> vrr; 38 for(int i=v.size()-1;i>0&&vrr.size()<m;--i) 39 { 40 if(vi[i]==-1) 41 vrr.emplace_back(i); 42 } 43 for(int i=v.size()-1;i>0&&vrr.size()<m;--i) 44 { 45 if(vi[i]==0) 46 vrr.emplace_back(i); 47 } 48 sort(vrr.begin(),vrr.end(),greater<int>()); 49 if(vr.empty()) 50 vr=vrr; 51 else 52 for(int i=0;i<vrr.size();++i) 53 { 54 if(vrr[i]>vr[i]) 55 { 56 vr=vrr; 57 break; 58 } 59 } 60 } 61 vi[fl]=0; 62 } 63 } 64 else 65 { 66 67 int vln=vl[n]; 68 int vn=abs(v[n]); 69 int vivn=vi[vn]; 70 71 if(v[n]>0) 72 { 73 if(vi[vn]<0) 74 { 75 vl[n]=-1; 76 dfs(n-1,ln+1,wn,un); 77 } 78 else if(vi[vn]>0) 79 { 80 vl[n]=1; 81 dfs(n-1,ln,wn,un); 82 } 83 else 84 { 85 vl[n]=1; 86 vi[vn]=1; 87 dfs(n-1,ln,wn,un-1); 88 vl[n]=-1; 89 vi[vn]=-1; 90 dfs(n-1,ln+1,wn+1,un-1); 91 } 92 } 93 else 94 { 95 if(vi[vn]<0) 96 { 97 vl[n]=1; 98 dfs(n-1,ln,wn,un); 99 } 100 else if(vi[vn]>0) 101 { 102 vl[n]=-1; 103 dfs(n-1,ln+1,wn,un); 104 } 105 else 106 { 107 vl[n]=1; 108 vi[vn]=-1; 109 dfs(n-1,ln,wn+1,un-1); 110 vl[n]=-1; 111 vi[vn]=1; 112 dfs(n-1,ln+1,wn,un-1); 113 } 114 } 115 vi[vn]=vivn; 116 vl[n]=vln; 117 return; 118 } 119 } 120 int main() 121 { 122 // freopen("data.txt","r",stdin); 123 ios::sync_with_stdio(false); 124 int x; 125 cin>>n>>m>>k; 126 vl.resize(n+1,0); 127 vi.resize(n+1,0); 128 v.push_back(0); 129 for(;n--;) 130 { 131 cin>>x; 132 v.push_back(x); 133 } 134 dfs(v.size()-1,0,0,v.size()-1); 135 136 if(vr.empty()) 137 cout<<"No Solution"; 138 else 139 { 140 cout<<vr[0]; 141 for(int i=1;i<vr.size();++i) 142 cout<<" "<<vr[i]; 143 } 144 return 0; 145 }