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题目链接：​​http://acm.hdu.edu.cn/showproblem.php?pid=4849​​


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Problem Description



Doge, tired of being a popular image on internet, is considering moving to another city for a new way of life.
   In his country there are N (2 ≤N≤ 1000) cities labeled 0 . . . N - 1. He is currently in city 0. Meanwhile, for each pair of cities, there exists a road connecting them, costing Ci,j (a positive integer) for traveling from city i to city j. Please note that Ci,j may not equal to Cj,i for any given i ≠ j.
   Doge is carefully examining the cities: in fact he will divide cities (his current city 0 is NOT included) into M (2 ≤ M ≤ 106) categories as follow:  If the minimal cost from his current city (labeled 0) to the city i is Di, city i belongs to category numbered Di mod M.Doge wants to know the “minimal” category (a category with minimal number) which contains at least one city.
   For example, for a country with 4 cities (labeled 0 . . . 3, note that city 0 is not considered), Doge wants to divide them into 3 categories.  Suppose category 0 contains no city, category 1 contains city 2 and 3, while category 2 contains city 1, Doge consider category 1 as the minimal one.
   Could you please help Doge solve this problem?


Note:

   Ci,j is generated in the following way:
   Given integers X0, X1, Y0, Y1, (1 ≤  X0, X1, Y0, Y1≤ 1234567), for k ≥ 2 we have
   Xk  =  (12345 + Xk-1 * 23456 + Xk-2 * 34567 + Xk-1 * Xk-2 * 45678)  mod  5837501
   Yk  =  (56789 + Yk-1 * 67890 + Yk-2 * 78901 + Yk-1 * Yk-2 * 89012)  mod  9860381
The for k ≥ 0 we have

                  Zk = (Xk * 90123 + Yk ) mod 8475871 + 1

Finally for 0 ≤ i, j ≤ N - 1 we have

                  Ci,j = Zi*n+j    for  i ≠ j
                  Ci,j = 0       for  i = j


 


Input



There are several test cases. Please process till EOF.
   For each test case, there is only one line containing 6 integers N,M,X0,X1,Y0,Y1.See the description for more details.


 


Output



For each test case, output a single line containing a single integer: the number of minimal category.


 


Sample Input



3 10 1 2 3 4
4 20 2 3 4 5


Sample Output



1
10

For the first test case, we have


     0     1     2     3     4     5     6     7     8
X    1     2   185180  788997 1483212 4659423 4123738 2178800  219267
Y    3     4  1633196 7845564 2071599 4562697 3523912 317737  1167849
Z  90127   180251 1620338 2064506  625135 5664774 5647950 8282552 4912390

the cost matrix C is
              0   180251 1620338
                 2064506    0  5664774 
                  5647950 8282552    0  

So the minimal cost from city 0 to city 1 is 180251, while the distance to city 2 is 1620338. Given M = 10, city 1 and city 2 belong to category 1 and 8 respectively. Since only category 1 and 8 contain at least one city, the minimal one of them, category 1, is the desired answer to Doge’s question.


题意：找从起点0開始到n-1各点的最短距离！注意要先模m后再找最短距离，也就是说存在原路径并非最短距离，可是模上m后就是最短的距离的情况！


dijkstra代码例如以下：

#include <cstdio>
#include <iostream>
#include <algorithm>
#define MAX 1017
#define INF 1000000000
using namespace std;
__int64 c[MAX][MAX];
__int64 x[MAX*MAX], y[MAX*MAX], z[MAX*MAX];

void dijkstra (__int64 mat[][MAX],int n, int s,int f)
{//s为起点。 f：为终点
    __int64 dis[MAX];//记录到不论什么点的最短距离
    __int64 mark[MAX];//记录被选中的结点 
    int i,j,k = 0;
    for(i = 0 ; i < n ; i++)//初始化全部结点。每一个结点都没有被选中 
        mark[i] = 0;
    for(i = 0 ; i < n ; i++)//将每一个结点到start结点weight记录为当前distance 
    {
        dis[i] = mat[s][i];
        //path[i] = s;
    }
    mark[s] = 1;//start结点被选中 
    //path[s] = 0;
    dis[s] = 0;//将start结点的的距离设置为0 
    __int64 min ;//设置最短的距离。 
    for(i = 1 ; i < n; i++)
    {
        min = INF;
        for(j = 0 ; j < n;j++)
        {
            if(mark[j] == 0  && dis[j] < min)//未被选中的结点中。距离最短的被选中 
            {
                min = dis[j] ;
                k = j;
            }
        }
        mark[k] = 1;//标记为被选中 
        for(j = 0 ; j < n ; j++)
        {
            if( mark[j] == 0  && (dis[j] > (dis[k] + mat[k][j])))//改动剩余结点的最短距离 
            {
                dis[j] = dis[k] + mat[k][j];
            }
        }
    } 
} 
int main()
{
    int n, m;
    int i, j;

    while(~scanf("%d%d",&n,&m))
    {
        scanf("%I64d%I64d%I64d%I64d",&x[0],&x[1],&y[0],&y[1]);
        for(i = 0; i < 2; i++)
        {
            z[i] = (x[i]*90123+y[i])%8475871+1;
        }
        for(i = 2; i <= n*n; i++)
        {
            x[i]=(12345+x[i-1]*23456+x[i-2]*34567+x[i-1]*x[i-2]*45678)%5837501;
            y[i]=(56789+y[i-1]*67890+y[i-2]*78901+y[i-1]*y[i-2]*89012)%9860381;
            z[i]=(x[i]*90123+y[i])%8475871+1;
        }
        for(i = 0; i < n; i++)
        {
            for(j = 0; j < n; j++)
            {
                if(i == j)
                    c[i][j] = 0;
                else
                    c[i][j] = z[i*n+j];
            }
        }
        __int64 mmax = INF, ans;
        dijkstra(c,n,0,n-1);
        for(i = 1; i < n; i++)
        {
            ans = dis[i];
            if(ans == 0)
                continue;
            ans%=m;
            if(mmax > ans)
            {
                mmax = ans;
            }
        }
        printf("%I64d\n",mmax);
    }
    return 0;
}