King

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 56 Accepted Submission(s): 30

Problem 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

Sample Output

lamentable kingdom
successful conspiracy

Source

Central Europe 1997

Recommend

/*
题意：现在有一个序列S={ a1,a2....an},问存不存在一个这样的序列 s ={s1,s2...sn} si={asi,asi+1,asi+2...asi+ni); 使得si大于或者小于ki

初步思路：这种题有不等号的，基本上就是差分约数了，定义数组T[i]表示S数组的前缀和，然后si=T[si+ni]-T[si-1]>(or<)ki,进行建边

#注意：spfa判环的时候，不再是n了，而是n+1因为Si的元素有n+1种
*/
#include<bits/stdc++.h>
using namespace std;
int n,m;
int si,ni,ki;
char str[3];
/*****************************************************spaf模板*****************************************************/
template<int N,int M>
struct Graph
{
    int top;
    struct Vertex{
        int head;
    }V[N];
    struct Edge{
        int v,next;
        int w;
    }E[M];
    void init(){
        memset(V,-1,sizeof(V));
        top = 0;
    }
    void add_edge(int u,int v,int w){
        E[top].v = v;
        E[top].w = w;
        E[top].next = V[u].head;
        V[u].head = top++;
    }
};

Graph<50005,150005> g;

const int N = 5e4 + 5;

int d[N];//从某一点到i的最短路
int inqCnt[N];

bool inq[N];//标记走过的点

bool spfa()
{
    memset(inqCnt,0,sizeof(inqCnt));
    memset(inq,false,sizeof(inq));
    memset(d,-63,sizeof(d));
    queue<int> Q;
    for(int i=0;i<=n;++i){
        inq[i] = 0; inqCnt[i] = 1;
        d[i] = 0; Q.push(i);
    }
    while(!Q.empty())
    {
        int u = Q.front();
        for(int i=g.V[u].head;~i;i=g.E[i].next)//遍历所有这个点相邻的点
        {
            int v = g.E[i].v;
            int w = g.E[i].w;
            if(d[u]+w>d[v])//进行放缩
            {
                d[v] = d[u] + w;
                if(!inq[v])//如果这个点没有遍历过
                {
                    Q.push(v);
                    inq[v] = true;
                    if(++inqCnt[v] > n+1)//n+1次才可能有负环，因为Si中有n+1个数
                        return true;
                }
            }
        }
        Q.pop();//将这个点出栈
        inq[u] = false;
    }
    return false;
}
/*****************************************************spaf模板*****************************************************/
int main(){
    // freopen("in.txt","r",stdin);
    while(scanf("%d",&n)!=EOF&&n){
        g.init();
        scanf("%d",&m);
        for(int i=0;i<m;i++){
            scanf("%d%d%s%d",&si,&ni,str,&ki);
            //T[si+ni]-T[si-1]>ki
            if(str[0]=='g') g.add_edge(si-1,si+ni,ki+1);
            //T[si+ni]-T[si-1]<ki
            else if(str[0]=='l') g.add_edge(si+ni,si-1,1-ki);
        }
        printf("%s
",spfa()? "successful conspiracy":"lamentable kingdom");  
    }
    return 0;
}