CodeForces 156B Suspects(枚举)
2016-05-07 07:53
323 查看
B. Suspects
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime.
To find out which one did it, the detective lines up the suspects and numbered them from 1 to n.
After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect
number ai",
or "Suspect number aididn't
commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand
this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n)
— the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th
line contains either "+ai"
(without the quotes), if the suspect number isays that the crime was committed by suspect number ai,
or "-ai"
(without the quotes), if the suspect number i says that the suspect number ai didn't
commit the crime (ai is
an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should
contain "Truth" if suspect number i has
told the truth for sure. Print "Lie" if the suspect number ilied
for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
input
output
input
output
input
output
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
As Sherlock Holmes was investigating a crime, he identified n suspects. He knows for sure that exactly one of them committed the crime.
To find out which one did it, the detective lines up the suspects and numbered them from 1 to n.
After that, he asked each one: "Which one committed the crime?". Suspect number i answered either "The crime was committed by suspect
number ai",
or "Suspect number aididn't
commit the crime". Also, the suspect could say so about himself (ai = i).
Sherlock Holmes understood for sure that exactly m answers were the truth and all other answers were a lie. Now help him understand
this: which suspect lied and which one told the truth?
Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 0 ≤ m ≤ n)
— the total number of suspects and the number of suspects who told the truth. Next n lines contain the suspects' answers. The i-th
line contains either "+ai"
(without the quotes), if the suspect number isays that the crime was committed by suspect number ai,
or "-ai"
(without the quotes), if the suspect number i says that the suspect number ai didn't
commit the crime (ai is
an integer, 1 ≤ ai ≤ n).
It is guaranteed that at least one suspect exists, such that if he committed the crime, then exactly m people told the truth.
Output
Print n lines. Line number i should
contain "Truth" if suspect number i has
told the truth for sure. Print "Lie" if the suspect number ilied
for sure and print "Not defined" if he could lie and could tell the truth, too, depending on who committed the crime.
Examples
input
1 1 +1
output
Truth
input
3 2 -1 -2 -3
output
Not defined Not defined Not defined
input
4 1 +2 -3 +4 -1
output
Lie Not defined Lie Not defined 枚举i是罪犯,然后看哪些人说了真话,人数等于m说明这个人可能是罪犯#include <iostream> #include <string.h> #include <stdlib.h> #include <algorithm> #include <math.h> #include <stdio.h> using namespace std; #define MAX 100000 int n,m; int a[MAX+5]; int b[MAX+5]; int f[MAX+5]; int tag[MAX+5]; int main() { scanf("%d%d",&n,&m); int x; memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); int sum=0; for(int i=1;i<=n;i++) { scanf("%d",&f[i]); if(f[i]>0) { a[f[i]]++; } else { b[-f[i]]++; sum++; } } memset(tag,0,sizeof(tag)); int k=0; for(int i=1;i<=n;i++) { if(a[i]+sum-b[i]==m) { tag[i]=1; k++; } } for(int i=1;i<=n;i++) { if(f[i]>0) { if(tag[f[i]]&&k==1) printf("Truth\n"); else if(!tag[f[i]]) printf("Lie\n"); else printf("Not defined\n"); } else { if(!tag[-f[i]]) printf("Truth\n"); else if(tag[-f[i]]&&k==1) printf("Lie\n"); else printf("Not defined\n"); } } return 0; }
相关文章推荐
- CodeForces 156B Suspects(枚举)
- CodeForces 156A Message(暴力)
- CodeForces 156A Message(暴力)
- 区域生长算法(附MATLAB代码实现)
- CodeForces 157B Trace
- CodeForces 157B Trace
- CodeForces 157A Game Outcome
- CodeForces 157A Game Outcome
- win7任务栏图标显示异常的解决方法
- 动态链接库DLL的加载:隐式加载(载入时加载)和显式加载(运行时加载)
- HDU 3578 Greedy Tino(双塔DP)
- 第一个DLL程序:动态链接库DLL教程,30分钟快速上手
- HDU 3578 Greedy Tino(双塔DP)
- POJ 2609 Ferry Loading(双塔DP)
- POJ 2609 Ferry Loading(双塔DP)
- 最小-最大计划报表-无法生成申请
- setContentView( )方法
- 数据科学家应该掌握的12种机器学习算法
- 吃了程序员这碗青春饭 40岁你该怎么办?
- setContentView( )方法