这题目乍眼一看还以为是2-sat。其实很水的，O(n)就解了。枚举每个人，假设其作为凶手。观察是否满足条件。
然后再对满足的数目分类讨论，进行求解。
/* 156B */
#include <iostream>
#include <string>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#include <deque>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <climits>
#include <cctype>
#include <cassert>
#include <functional>
#include <iterator>
#include <iomanip>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,1024000") #define sti set<int>
#define stpii set<pair<int, int> >
#define mpii map<int,int>
#define vi vector<int>
#define pii pair<int,int>
#define vpii vector<pair<int,int> >
#define rep(i, a, n) for (int i=a;i<n;++i)
#define per(i, a, n) for (int i=n-1;i>=a;--i)
#define clr clear
#define pb push_back
#define mp make_pair
#define fir first
#define sec second
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define lson l, mid, rt<<1
#define rson mid+1, r, rt<<1|1 const int maxn = 1e5+;
int pos[maxn], neg[maxn];
vi P[maxn];
vi N[maxn];
int ans[maxn];
int st[maxn];
char s[][] = {
"Not defined",
"Truth",
"Lie"
}; int main() {
ios::sync_with_stdio(false);
#ifndef ONLINE_JUDGE
freopen("data.in", "r", stdin);
freopen("data.out", "w", stdout);
#endif int n, m, x;
int pn = , nn = ; scanf("%d %d", &n, &m);
rep(i, , n+) {
scanf("%d", &x);
if (x > ) {
++pos[x];
P[x].pb(i);
++pn;
} else {
x = -x;
++neg[x];
N[x].pb(i);
++nn;
}
} int tmp;
vi vc; rep(i, , n+) {
// i is murder
tmp = pos[i] + nn - neg[i];
if (tmp == m)
vc.pb(i);
} int sz = SZ(vc);
if (sz == ) {
x = vc[];
rep(i, , n+)
st[i] = -;
st[x] = ;
} else {
rep(i, , n+)
st[i] = -;
rep(i, , sz)
st[vc[i]] = ;
}
rep(i, , n+) {
sz = SZ(P[i]);
if (sz && st[i]) {
x = st[i]> ? :;
rep(j, , sz)
ans[P[i][j]] = x;
}
sz = SZ(N[i]);
if (sz && st[i]) {
x = st[i]< ? :;
rep(j, , sz)
ans[N[i][j]] = x;
}
} rep(i, , n+)
puts(s[ans[i]]); #ifndef ONLINE_JUDGE
printf("time = %d.\n", (int)clock());
#endif return ;
}