POJ 1364 King 差分约束系统

/**
 * @file main.cpp
 * @brief 差分约束系统,题目意思为:
 *        国王给出m组假设,计算满足所有假设的数组是否存在
 *        假设分两种:
 *        1. 从si开始的连续ni+1个数字相加小于ki
 *        2. 从si开始的连续ni+1个数字相加大于ki
 *
 *        令s[i]表示a[0]+a[1]+...+a[i],则以上条件可以描述为:
 *        1. s[si + ni] - s[si - 1] < ki
 *        2. s[si + ni] - s[si - 1] > ki
 *
 *        即(可以把大于等于化为大于,通过整数运算)
 *        1. if (s[si + ni] + (1 - ki) > s[si - 1])
 *            s[si - 1] = s[si + ni] + (1 - ki)
 *        2. if (s[si - 1] + (1 + ki) > s[si + ni])
 *            s[si + ni] = s[si - 1] + (1 + ki)
 *
 *        构图:
 *        1. (si + ni, si - 1)添加权值1-ki
 *        2. (si - 1, si + ni)添加权值1+ki
 *
 *        使用bellman_ford检测是否有负环即可
 *
 * @author yekeren
 * @version 1.0.0
 * @date 2013-06-07
 */

#include 
#include 

#define LIMIT 101

struct edge_t {
    int a, b, c;
} k[LIMIT];

bool bellman_ford(int n, int m)
{
    int dist[LIMIT] = { 0 };

    for (int t = 0; t < n - 1; ++t)
    {
        for (int i = 0; i < m; ++i)
        {
            if (dist[k[i].a] + k[i].c > dist[k[i].b]) {
                dist[k[i].b] = dist[k[i].a] + k[i].c;
            }
        }
    }

    //
    //已经松弛n-1次,看能否再松弛
    //
    bool flag = false;
    for (int i = 0; i < m; ++i)
    {
        if (dist[k[i].a] + k[i].c > dist[k[i].b])
        {
            dist[k[i].b] = dist[k[i].a] + k[i].c;
            return false;
        }
    }

    return true;
}

int main(int argc, char *argv[])
{
    int n, m;

    while (std::cin >> n, n != 0)
    {
        std::cin >> m;

        for (int i = 0; i < m; ++i)
        {
            char oi[4];
            int si, ni, ki;

            std::cin >> si >> ni >> oi >> ki;

            if (strcmp(oi, "gt") == 0) 
            {
                k[i].a = si - 1;
                k[i].b = si + ni;
                k[i].c = 1 + ki;
            } 
            else if (strcmp(oi, "lt") == 0) 
            {
                k[i].a = si + ni;
                k[i].b = si - 1;
                k[i].c = 1 - ki;
            }
        }

        if (bellman_ford(n + 1, m)) {
            std::cout << "lamentable kingdom" << std::endl;
        } else {
            std::cout << "successful conspiracy" << std::endl;
        }
    }

    return 0;
}