hdu4849 最短路

题意：

      让你求0到所有点最短路中对m取余最小的那个数。

思路：

      简单题，直接根据题目给的公式把z求出来，然后建边，然后最短路，然后枚举每一个点对m取余记录最小，然后输出答案，然后ac.

#include<stdio.h>
#include<queue>

#define N_node 1111
#define N_edge 1111111
#define INF 1000000000

using namespace std;

typedef struct
{
   int to ,next;
   __int64 cost;
}STAR;

STAR E[N_edge];
int list[N_node] ,tot;
__int64 s_x[N_node];
__int64 Z[N_edge];

void add(int a ,int b ,__int64 c)
{
   E[++tot].to = b;
   E[tot].cost = c;
   E[tot].next = list[a];
   list[a] = tot;
}

void spfa(int s ,int n)
{
   int mark[N_node] = {0};
   for(int i = 0 ;i <= n ;i ++)
   s_x[i] = INF;
   mark[s] = 1 ,s_x[s] = 0;
   queue<int>q;
   q.push(s);
   while(!q.empty())
   {
      int xin ,tou;
      tou = q.front();
      q.pop();
      mark[tou] = 0;
      for(int k = list[tou] ;k ;k = E[k].next)
      {
         xin = E[k].to;
         if(s_x[xin] > s_x[tou] + E[k].cost)
         {
            s_x[xin] = s_x[tou] + E[k].cost;
            if(!mark[xin])
            {
               mark[xin] = 1;
               q.push(xin);
            }
         }
      }
   }
   return ;
}

int main ()
{
   int n ,i ,j;
   __int64 x1 ,x2 ,y1 ,y2 ,x ,y ,m;
   while(~scanf("%d %I64d %I64d %I64d %I64d %I64d" ,&n ,&m ,&x1 ,&x2 ,&y1 ,&y2))
   {
      Z[1] = (x1 * 90123 + y1) % 8475871 + 1;
      Z[2] = (x2 * 90123 + y2) % 8475871 + 1;
      for(i = 3 ;i <= n * n ;i ++)
      {
         x  = (12345 + x2 * 23456 + x1 * 34567 + x2 * x1 * 45678)  %  5837501;
         y  = (56789 + y2 * 67890 + y1 * 78901 + y2 * y1 * 89012)  %  9860381;
         Z[i] = (x * 90123 + y) % 8475871 + 1;
         x1 = x2 ,y1 = y2;
         x2 = x ,y2 = y;
      }

      memset(list ,0 ,sizeof(list)) ,tot = 1;
      for(i = 1 ;i <= n ;i ++)
      for(j = 1 ;j <= n ;j ++)
      if(i == j) add(i ,j ,0);
      else add(i ,j ,Z[(i - 1) * n + j]);
      spfa(1 ,n);
      __int64 min = INF;
      for(i = 2 ;i <= n ;i ++)
      if(min > s_x[i] % m)
      min = s_x[i] % m;
      printf("%I64d\n" ,min);
   }
   return 0;
}