Wow! Such City!

Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 102400/102400 K (Java/Others)

Total Submission(s): 824    Accepted Submission(s): 310

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 C

_i,

_j (a positive integer) for traveling from city i to city j. Please note that C

_i,

_j may not equal to C

_j,

_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 ≤ 10

⁶) 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 

D_i 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:

C_i,_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 + X_k-1 * 23456 + X_k-2 * 34567 + X_k-1 * X_k-2 * 45678)  mod  5837501

Yk  = (56789 + Y_k-1 * 67890 + Y_k-2 * 78901 + Y_k-1 * Y_k-2 * 89012)  mod  9860381

The for k ≥ 0 we have

Z_k = (X_k * 90123 + Y_k ) mod 8475871 + 1

Finally for 0 ≤ i, j ≤ N - 1 we have

C_i,_j = Z_i*n+j for i ≠ j

C_i,_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,X

₀,X

₁,Y

₀,Y

₁.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

头一次遇到区域赛的最短路的题目。。

尽管不是纯最短路（只是也差点儿相同了。。）权值由递推公式（题目中已给出）生成，然后跑一遍dijkstra，起点为1，求dis[i]%m的最小值。权值注意超出int范围要用lld

#include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #define inf 0x3f3f3f3f#define ll long longusing namespace std;ll dis[1010],v[1010];ll map[1010][1010];int x0,x1,y0,y1,n,m;void dijkstra(){    int minx,k=0;    for(int i=0; i<=n; i++)    {        dis[i]=map[0][i];;        v[i]=0;    }    dis[0]=0;    for(int j=0; j
            
             dis[i]) { minx=dis[i]; k=i; } } v[k]=1; for(int i=0; i
             
              dis[k]+map[k][i]) { dis[i]=dis[k]+map[k][i]; } } } return ;}ll xx[1002000],yy[1002000],zz[1002000];int main(){ while(scanf("%d%d%d%d%d%d",&n,&m,&x0,&x1,&y0,&y1)!=EOF) { for(int i=0; i