hdu4849 Wow! Such City!(最短路dijkstra)

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题目链接：

pid=4849">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;
}