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.
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 y[%d]=%lld z[%d]=%lld ",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 ",c[i][j]); 32 } 33 //printf(" "); 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 ",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 ",i); 77 break; 78 } 79 } 80 } 81 }