HDU 4255 a famous grid-analog-sieve prime number-BFS

# Include <stdio. h> # include <string. h ># include <queue> using namespace STD; int S, T; struct node {int X, Y;} Shu [40001]; node SS, TT; int Suy [40001]; int map [201] [201]; int Fang [201]; void inif () // calculates the number of workers {for (INT I = 1; I <= 200; I ++) Fang [I] = I * I;} void Xu (int I) // calculates the position of the digit I {int L, R; for (r = 2; r <= 200; ++ R) if (I <Fang [R]) break; L = R-1; If (L % 2) {If (I-L * l) <= (R * r-l * l)/2) {Shu [I]. X = Shu [L * l]. x-(I-L * l) + 1; Shu [I]. y = Shu [L * l]. Y + 1 ;} Else {Shu [I]. X = Shu [R * r]. x; Shu [I]. y = Shu [R * r]. Y + R * r-I;} else {If (I-L * l) <= (R * r-l * l)/2) {Shu [I]. X = Shu [L * l]. X + I-l * l-1; Shu [I]. y = Shu [L * l]. y-1;} else {Shu [I]. X = Shu [R * r]. x; Shu [I]. y = Shu [R * r]. y-(R * r-I) ;}} void INI () // initialize {inif (); int I, Ju; Shu [1]. X = 101; Shu [1]. y= 100; Shu [4]. X = 100; Shu [4]. y = 100; for (I = 3; I <= 200; I ++) {if (I % 2) {ju = (I-1)/2; shu [I * I]. X = Shu [1]. X + Ju; Shu [I * I]. y = Shu [1]. Y + Ju;} else {ju = (I-2)/2; Shu [I * I]. X = Shu [4]. x-ju; Shu [I * I]. y = Shu [4]. y-ju ;}for (I = 2; I <= 40000; ++ I) if (Shu [I]. X = 0) XU (I);} void SU () // screen prime number {int I, j; memset (Suy, 0, sizeof (Suy )); suy [1] = 1; for (I = 2; I <= 200; I ++) {If (Suy [I] = 0) {for (j = I + I; j <= 40000; j = J + I) Suy [J] = 1; // The number is 1} memset (MAP, 0, sizeof (MAP);} struct qnode {int X, Y, N ;}; int V [200] [200]; int dir [] [2] =, 0, 1, 0, 0,-1}; int DFS () // find the shortest path {int K; queue <qnode> q; qnode Cur, next; cur. X = ss. x; cur. y = ss. y; cur. n = 0; memset (v, 0, sizeof (v); V [cur. x] [cur. y] = 1; q. push (cur); While (! Q. empty () {cur = Q. front (); q. pop (); For (k = 0; k <4; k ++) {next. X = cur. X + dir [k] [0]; next. y = cur. Y + dir [k] [1]; next. N = cur. n + 1; if (next. X = TT. X & next. y = TT. y) return next. n; If (next. x> = 1 & next. X <=200 & next. y> = 1 & next. Y <= 200 & map [next. x] [next. y] = 0 & V [next. x] [next. y] = 0) {v [next. x] [next. y] = 1; q. push (next) ;}}return-1;} void biaosu () // mark the prime number {int I, X, Y; for (I = 1; I <= 40000; ++ I) {If (! Suy [I]) {x = Shu [I]. x; y = Shu [I]. y; Map [x] [Y] = 1 ;}} int main () {int ret, CAS = 1; INI (); SU (); biaosu (); while (scanf ("% d", & S, & T )! = EOF) {Ss = Shu [s]; TT = Shu [T]; ret = DFS (); If (ret =-1) printf ("case % d: impossible \ n ", CAS ++); else printf (" case % d: % d \ n ", CAS ++, RET);} return 0 ;}