hdu4849 Wow! Such City!

hdu4849 Wow! Such City!(最短路dijkstra)

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  9860381The for k ≥ 0 we have                  Zk = (Xk * 90123 + Yk ) mod 8475871 + 1Finally 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 44 20 2 3 4 5

Sample Output

110For the first test case, we have   0   1   2   3   4   5   6   7   8X   1   2 185180 7889971483212465942341237382178800 219267Y   3   4163319678455642071599456269735239123177371167849Z 90127 18025116203382064506 6251355664774564795082825524912390the cost matrix C is            0 1802511620338                2064506   05664774                56479508282552   0So 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开始到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不是最短距离，但是模上m后就是最短的距离的情况！

dijkstra代码如下：

[cpp] view plaincopyprint?

1. #include <cstdio>  
2. #include <iostream>  
3. #include <algorithm>  
4. #define MAX 1017  
5. #define INF 1000000000  
6. using namespace std;  
7. __int64 c[MAX][MAX];  
8. __int64 x[MAX*MAX], y[MAX*MAX], z[MAX*MAX];  
9.   
10. void dijkstra (__int64 mat[][MAX],int n, int s,int f)  
11. {//s为起点， f：为终点  
12.     __int64 dis[MAX];//记录到任何点的最短距离  
13.     __int64 mark[MAX];//记录被选中的结点   
14.     int i,j,k = 0;  
15.     for(i = 0 ; i < n ; i++)//初始化所有结点，每个结点都没有被选中   
16.         mark[i] = 0;  
17.     for(i = 0 ; i < n ; i++)//将每个结点到start结点weight记录为当前distance   
18.     {  
19.         dis[i] = mat[s][i];  
20.         //path[i] = s;  
21.     }  
22.     mark[s] = 1;//start结点被选中   
23.     //path[s] = 0;  
24.     dis[s] = 0;//将start结点的的距离设置为0   
25.     __int64 min ;//设置最短的距离。   
26.     for(i = 1 ; i < n; i++)  
27.     {  
28.         min = INF;  
29.         for(j = 0 ; j < n;j++)  
30.         {  
31.             if(mark[j] == 0  && dis[j] < min)//未被选中的结点中，距离最短的被选中   
32.             {  
33.                 min = dis[j] ;  
34.                 k = j;  
35.             }  
36.         }  
37.         mark[k] = 1;//标记为被选中   
38.         for(j = 0 ; j < n ; j++)  
39.         {  
40.             if( mark[j] == 0  && (dis[j] > (dis[k] + mat[k][j])))//修改剩余结点的最短距离   
41.             {  
42.                 dis[j] = dis[k] + mat[k][j];  
43.             }  
44.         }  
45.     }   
46. }   
47. int main()  
48. {  
49.     int n, m;  
50.     int i, j;  
51.       
52.     while(~scanf("%d%d",&n,&m))  
53.     {  
54.         scanf("%I64d%I64d%I64d%I64d",&x[0],&x[1],&y[0],&y[1]);  
55.         for(i = 0; i < 2; i++)  
56.         {  
57.             z[i] = (x[i]*90123+y[i])%8475871+1;  
58.         }  
59.         for(i = 2; i <= n*n; i++)  
60.         {  
61.             x[i]=(12345+x[i-1]*23456+x[i-2]*34567+x[i-1]*x[i-2]*45678)%5837501;  
62.             y[i]=(56789+y[i-1]*67890+y[i-2]*78901+y[i-1]*y[i-2]*89012)%9860381;  
63.             z[i]=(x[i]*90123+y[i])%8475871+1;  
64.         }  
65.         for(i = 0; i < n; i++)  
66.         {  
67.             for(j = 0; j < n; j++)  
68.             {  
69.                 if(i == j)  
70.                     c[i][j] = 0;  
71.                 else  
72.                     c[i][j] = z[i*n+j];  
73.             }  
74.         }  
75.         __int64 mmax = INF, ans;  
76.         dijkstra(c,n,0,n-1);  
77.         for(i = 1; i < n; i++)  
78.         {  
79.             ans = dis[i];  
80.             if(ans == 0)  
81.                 continue;  
82.             ans%=m;  
83.             if(mmax > ans)  
84.             {  
85.                 mmax = ans;  
86.             }  
87.         }  
88.         printf("%I64d\n",mmax);  
89.     }  
90.     return 0;  
91. }