POJ 1346--King 【差分约束】

King

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 10994 Accepted: 4025

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 21 2 gt 02 2 lt 21 21 0 gt 01 0 lt 00

Sample Output

lamentable kingdomsuccessful conspiracy

题意：一个长度为n的序列和它的m个子序列，每个子序列的和都有一个限制k，gt代表大于k，lt代表小于k。 问你这样的序列是否存在。

思路：大于k == 大于等于k+1, 小于k == 小于等于k-1。 图可能是不连通的，所以需要建立虚拟源点并和所有顶点连边来保证图的连通，接下来就是判断是否存在负环就可以了。

注意：图中有 0 到 n 共 n + 1 个点。

最长路写法：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#define maxn 1100#define INF 0x3f3f3f3fusing namespace std;struct node {    int u ,v, w, next;};node edge[1100000];int head[maxn], dist[maxn], used[maxn],cnt;bool vis[maxn];int N, M;void init(){    cnt = 0;    memset(head, -1, sizeof(head));}void add(int u, int v, int w){    edge[cnt] = {u, v, w, head[u]};    head[u] = cnt++;}void getmap(){    int a, b, k;    char str[10];    while(M--){        scanf("%d%d%s%d", &a, &b, str, &k);        if(str[0] == 'g')            add(a - 1, a + b, k + 1);        else if(str[0] == 'l')            add(a + b, a - 1, 1 - k);    }    for(int i = 0; i <= N; ++i)// N + 1为虚拟源点        add(N + 1, i, 0);}int SPFA(){    for(int i = 0;i <= N + 1; ++i){        dist[i] = -INF;        vis[i] = 0;        used[i] = 0;    }    used[N + 1] = 1;    vis[N + 1] = 1;    dist[N + 1] = 0;    queue<int>q;    q.push(N + 1);    while(!q.empty()){        int u = q.front();        q.pop();        vis[u] = 0;        for(int i = head[u]; i != -1; i = edge[i].next){            int v = edge[i].v;            int w = edge[i].w;            if(dist[v] < dist[u] + w){                dist[v] = dist[u] + w;                if(!vis[v]){                    vis[v] = 1;                    used[v]++;                    if(used[v] > N + 1 )                        return 0;                    q.push(v);                }            }        }    }    return 1;}int main (){    while(scanf("%d", &N), N){        scanf("%d", &M);        init();        getmap();        if(SPFA())            printf("lamentable kingdom\n");        else            printf("successful conspiracy\n");    }    return 0;}