King

http://162.105.81.212/JudgeOnline/problem?id=1364

Description

Once, in one kingdom, there was a queen and that queen was expecting a baby. The queen prayed: ``If my child was a son and if only he was a sound king.'' After nine months her child was born, and indeed, she gave birth to a nice son.
Unfortunately, as it used to happen in royal families, the son was a little retarded. After many years of study he was able just to add integer numbers and to compare whether the result is greater or less than a given integer number. In addition, the numbers had to be written in a sequence and he was able to sum just continuous subsequences of the sequence.

The old king was very unhappy of his son. But he was ready to make everything to enable his son to govern the kingdom after his death. With regards to his son's skills he decided that every problem the king had to decide about had to be presented in a form of a finite sequence of integer numbers and the decision about it would be done by stating an integer constraint (i.e. an upper or lower limit) for the sum of that sequence. In this way there was at least some hope that his son would be able to make some decisions.

After the old king died, the young king began to reign. But very soon, a lot of people became very unsatisfied with his decisions and decided to dethrone him. They tried to do it by proving that his decisions were wrong.

Therefore some conspirators presented to the young king a set of problems that he had to decide about. The set of problems was in the form of subsequences Si = {aSi, aSi+1, ..., aSi+ni} of a sequence S = {a1, a2, ..., an}. The king thought a minute and then decided, i.e. he set for the sum aSi + aSi+1 + ... + aSi+ni of each subsequence Si an integer constraint ki (i.e. aSi + aSi+1 + ... + aSi+ni < ki or aSi + aSi+1 + ... + aSi+ni > ki resp.) and declared these constraints as his decisions.

After a while he realized that some of his decisions were wrong. He could not revoke the declared constraints but trying to save himself he decided to fake the sequence that he was given. He ordered to his advisors to find such a sequence S that would satisfy the constraints he set. Help the advisors of the king and write a program that decides whether such a sequence exists or not.

Input

The input consists of blocks of lines. Each block except the last corresponds to one set of problems and king's decisions about them. In the first line of the block there are integers n, and m where 0 < n <= 100 is length of the sequence S and 0 < m <= 100 is the number of subsequences Si. Next m lines contain particular decisions coded in the form of quadruples si, ni, oi, ki, where oi represents operator > (coded as gt) or operator < (coded as lt) respectively. The symbols si, ni and ki have the meaning described above. The last block consists of just one line containing 0.

Output

The output contains the lines corresponding to the blocks in the input. A line contains text successful conspiracy when such a sequence does not exist. Otherwise it contains text lamentable kingdom. There is no line in the output corresponding to the last ``null'' block of the input.

Sample Input

4 2

1 2 gt 0

2 2 lt 2

1 2

1 0 gt 0

1 0 lt 0

0

Sample Output

lamentable kingdom

successful conspiracy

Source

Central Europe 1997

又是WA了n次的题目，主要错误有三个： 一是构图时大于，小于搞反了，明知道这里可能会出错，还是不愿先用在纸上模拟一下,

S[b] <= S[a] + k , 在图中加一条由a指向b权值为k的单向边.二是不能想解决Interval时只用添加最后一个节点入队列就可以了，

要添加所有的节点入队列，三是不是n个节点，是n+1个节点，所以在判断有负环应该大于n而不是大于等于n.

学差分约束系统，核心就是学如何构图，如何发现它可以用图来表示。

目前碰到的题目主要特征是 ，有 区间，大于小于，至少，至多 ， 这几个关键字，构图时用Sigma(b) - Sigma(a)表示区间，

把区间内的各个点变成顶点.

#include <iostream> #include <queue> using namespace std ; struct Graph { int vex , dis , next ; } ; Graph G[200] ; int first[200] , l , r , dis[200] , covered[200] , n ; bool visited[200] ; void Readin () { int i , m , a , b , k , j ; char sign[4] ; scanf ("%d" , &m) ; l = 200 ; memset (first , -1 , sizeof (first)) ; for (i = r = 0 ; i < m ; i ++) { scanf ("%d%d%s%d" , &a , &b , sign , &k) ; b += a ; a -- ; if (l > a) l = a ; if (r < b) r = b ; if (sign[0] == 'l') { k-- ; G[i].dis = k ; G[i].vex = b ; G[i].next = first[a] ; first[a] = i ; } else { k ++ ; G[i].dis = -k ; G[i].vex = a ; G[i].next = first[b] ; first[b] = i ; } } for (r ++ , j = l ; j < r ; j ++ , i ++) { G[i].dis = 0 ; G[i].vex = j ; G[i].next = first[r] ; first[r] = i ; } } bool SPFA () { queue <int> Q ; int v , tmp ; memset (covered , 0 , sizeof (covered)) ; memset (visited , false , sizeof (visited)) ; dis[r] = 0 , covered[r] = 1 ; visited[r] = true ; Q.push (r) ; while (!Q.empty ()) { v = Q.front () ; Q.pop () ; visited[v] = false ; for (tmp = first[v] ; tmp != -1 ; tmp = G[tmp].next) if (!covered[G[tmp].vex] || dis[v] + G[tmp].dis < dis[G[tmp].vex]) { dis[G[tmp].vex] = dis[v] + G[tmp].dis ; if (covered[v] > n + 2) return false ; covered[G[tmp].vex] = covered[v] + 1 ; if (!visited[G[tmp].vex]) { visited[G[tmp].vex] = true ; Q.push (G[tmp].vex) ; } } } return true ; } int main () { while (scanf ("%d" , &n) , n) { Readin () ; if (SPFA ()) printf ("lamentable kingdom/n") ; else printf ("successful conspiracy/n") ; } return 0 ; }