HDU 4849 – Wow! Such City!

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

 InputThere 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. OutputFor each test case, output a single line containing a
single integer: the number of minimal category. Sample Input3 10 1 2 3 44 20 2 3 4 5 Sample Output110  Hint:

怎么说吧，就是一个又臭又长又水的单源最短路径外加算算MOD的题。

 #include<cstdio>
 #include<queue>
 #include<cstring>
 #define MAXN 1003
 #define MAXK 1000*1000+1000
 #define INF 0x3f3f3f3f
 using namespace std;
 typedef long long ll;
 int n,m;
 ll x[MAXK],y[MAXK],z[MAXK],c[MAXN][MAXN];
 int cate[+];
 void calc()
 {
     int max_k=(n-)*n+(n-);
     for(int k=;k<=max_k;k++)
     {
         if(k>=)
         {
             x[k]=( + (x[k-] * ) %  + (x[k-] * ) %  + (x[k-] * x[k-] * ) %  ) % ;
             y[k]=( + (y[k-] * ) %  + (y[k-] * ) %  + (y[k-] * y[k-] * ) %  ) % ;
         }
         z[k]=(x[k] *  + y[k] ) %  + ;
     }
     //for(int k=0;k<=max_k;k++) printf("x[%d]=%lld \t y[%d]=%lld \t z[%d]=%lld \n",k,x[k],k,y[k],k,z[k]);
     for(int i=;i<n;i++)
     {
         for(int j=;j<n;j++)
         {
             if(i==j) c[i][j]=;
             else c[i][j]=z[(i*n+j)];
             //printf("%lld\t",c[i][j]);
         }
         //printf("\n");
     }
 }
 bool vis[MAXN];
 ll d[MAXN];
 void spfa()
 {
     for(int i=;i<n;i++){
         vis[i]=;
         d[i]=INF;
     }
     vis[]=;
     d[]=;
     queue<int> q;
     q.push();
     while(!q.empty())
     {
         int u=q.front();q.pop();vis[u]=;
         for(int v=;v<n;v++)
         {
             if(u==v) continue;
             ll tmp=d[v];
             if(d[v]>d[u]+c[u][v]) d[v]=d[u]+c[u][v];
             if(d[v]<tmp && !vis[v]) q.push(v),vis[v]=;
         }
     }
 }
 int main()
 {
     while(scanf("%d %d %lld %lld %lld %lld",&n,&m,&x[],&x[],&y[],&y[])!=EOF)
     {
         calc();
         spfa();
         memset(cate,,sizeof(cate));
         for(int i=;i<n;i++)
         {
             //printf("d[%d]=%lld\n",i,d[i]);
             cate[(d[i]%m)]++;
         }
         for(int i=;i<m;i++)
         {
             if(cate[i]!=)
             {
                 printf("%d\n",i);
                 break;
             }
         }
     }
 }