HDU 4849 - Wow! Such City!

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

 
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
 
 
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。

 1 #include<cstdio>
 2 #include<queue>
 3 #include<cstring>
 4 #define MAXN 1003
 5 #define MAXK 1000*1000+1000
 6 #define INF 0x3f3f3f3f
 7 using namespace std;
 8 typedef long long ll;
 9 int n,m;
10 ll x[MAXK],y[MAXK],z[MAXK],c[MAXN][MAXN];
11 int cate[1000000+3];
12 void calc()
13 {
14     int max_k=(n-1)*n+(n-2);
15     for(int k=0;k<=max_k;k++)
16     {
17         if(k>=2)
18         {
19             x[k]=(12345 + (x[k-1] * 23456) % 5837501 + (x[k-2] * 34567) % 5837501 + (x[k-1] * x[k-2] * 45678) % 5837501 ) % 5837501;
20             y[k]=(56789 + (y[k-1] * 67890) % 9860381 + (y[k-2] * 78901) % 9860381 + (y[k-1] * y[k-2] * 89012) % 9860381 ) % 9860381;
21         }
22         z[k]=(x[k] * 90123 + y[k] ) % 8475871 + 1;
23     }
24     //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]);
25     for(int i=0;i<n;i++)
26     {
27         for(int j=0;j<n;j++)
28         {
29             if(i==j) c[i][j]=0;
30             else c[i][j]=z[(i*n+j)];
31             //printf("%lld\t",c[i][j]);
32         }
33         //printf("\n");
34     }
35 }
36 bool vis[MAXN];
37 ll d[MAXN];
38 void spfa()
39 {
40     for(int i=1;i<n;i++){
41         vis[i]=0;
42         d[i]=INF;
43     }
44     vis[0]=1;
45     d[0]=0;
46     queue<int> q;
47     q.push(0);
48     while(!q.empty())
49     {
50         int u=q.front();q.pop();vis[u]=0;
51         for(int v=0;v<n;v++)
52         {
53             if(u==v) continue;
54             ll tmp=d[v];
55             if(d[v]>d[u]+c[u][v]) d[v]=d[u]+c[u][v];
56             if(d[v]<tmp && !vis[v]) q.push(v),vis[v]=1;
57         }
58     }
59 }
60 int main()
61 {
62     while(scanf("%d %d %lld %lld %lld %lld",&n,&m,&x[0],&x[1],&y[0],&y[1])!=EOF)
63     {
64         calc();
65         spfa();
66         memset(cate,0,sizeof(cate));
67         for(int i=1;i<n;i++)
68         {
69             //printf("d[%d]=%lld\n",i,d[i]);
70             cate[(d[i]%m)]++;
71         }
72         for(int i=0;i<m;i++)
73         {
74             if(cate[i]!=0)
75             {
76                 printf("%d\n",i);
77                 break;
78             }
79         }
80     }
81 }

 

posted @ 2017-05-05 23:30  Dilthey  阅读(277)  评论(0)  编辑  收藏  举报