HDU 1401 zoj1505 PKU 1198 bidirectional wide search

/* <Br/> I am afraid that I will never write it... <Br/> it is always strange to store the status, but the status in bfs dfs is very important. <br/> I wanted to write this question before... <Br/> the AC is finished .. <Br/> */<br/> # include <iostream> <br/> # include <cstdio> <br/> # include <bitset> <br/> # include <cstdio> <br/> # include <cstring> <br/> # include <algorithm> <br/> # include <queue> <br/> using namespace STD; <br/> const int n = (1 <24) + 10); <br/> bitset <n> hash1; // This can save memory <br/> bitset <n> hash2; <br/> int dir1 [4] [2] ={{-1, 0 }, {1, 0}, {0, 1}, {0,-1 }}; <br/> int dir2 [4] [2] = {-2, 0 },{ 2, 0 },{ 0, 2}, {0,-2 }}; <br/> char map1 [9] [9], MAP2 [9] [9]; <br/> struct dir <br/>{< br/> int X, Y; <br/> }; <br/> struct node <br/> {<br/> dir [4]; <br/> int step; <br/> char map [9] [9]; <br/> node () <br/> {<br/> for (INT I = 0; I <9; I ++) <br/> for (Int J = 0; j <9; j ++) <br/> map [I] [J] = 'a '; <br/> step = 0; <br/>}< br/>}; <br/> node start and end; <br/> inline bool CMP (const dir & A, const dir & B) <br/> {// sort coordinates to avoid unnecessary incoming queues <br /> If (. X = B. x) <br/> return. Y <B. y; <br/> else <br/> return. x <B. x; <br/>}< br/> inline int Hash (const node & node) <br/> {// convert coordinates to one-dimensional <br/> int num; <br/> int sum = 0; <br/> for (INT I = 0; I <4; I ++) <br/>{< br/> num = (node. dir [I]. x-1) * 8 + node. dir [I]. y; <br/> sum = 64 * sum + num; <br/>}< br/> return sum; <br/>}< br/> void BFS () <br/>{< br/> queue <node> q1, Q2; <br/> q1.push (start); <br/> q2.pus H (end); <br/> node now, next; <br/> hash1.reset (); <br/> hash2.reset (); <br/> hash1 [Hash (start)] = true; <br/> hash2 [Hash (end)] = true; <br/> int X, Y; <br/> for (INT step = 0; Step <4; Step ++) <br/>{< br/> while (! Q1.empty () <br/>{< br/> next = now = q1.front (); <br/> q1.pop (); <br/> If (next. step> 4) <br/> break; <br/> for (INT I = 0; I <4; I ++) // four dots D indicate the pawns, a Indicates blank <br/>{< br/> for (Int J = 0; j <4; j ++) <br/> {// first search for the direction <br/> next = now; <br/> X = now. dir [I]. X + dir1 [J] [0]; <br/> Y = now. dir [I]. Y + dir1 [J] [1]; <br/> If (x> = 1 & x <= 8 & Y> = 1 & Y <= 8 & now. map [x] [Y]! = 'D') <br/>{< br/> next. dir [I]. X = x; <br/> next. dir [I]. y = y; <br/> next. step = now. step + 1; <br/> next. map [x] [Y] = 'D'; <br/> next. map [now. dir [I]. x] [now. dir [I]. y] = 'a'; <br/> sort (next. dir, next. dir + 4, CMP); <br/> If (hash1 [Hash (next)]) <br/> continue; <br/> If (hash2 [Hash (next)]) <br/> {// if this state has been hashed by end, it indicates that printf ("Yes/N") exists. <br/> return; <br/>}< br/> hash1 [Hash (next)] = true; <br/> Q1. Push (next); <br/>}< br/> for (Int J = 0; j <4; j ++) <br/>{< br/> next = now; <br/> X = now. dir [I]. X + dir2 [J] [0]; <br/> Y = now. dir [I]. Y + dir2 [J] [1]; <br/> If (x> = 1 & x <= 8 & Y> = 1 & Y <= 8 & <br/> now. map [now. dir [I]. X + dir1 [I] [0] [now. dir [I]. Y + dir1 [I] [1] = 'D' <br/> & now. map [x] [Y]! = 'D') <br/>{< br/> next. dir [I]. X = x; <br/> next. dir [I]. y = y; <br/> next. step = now. step + 1; <br/> next. map [x] [Y] = 'D'; <br/> next. map [now. dir [I]. x] [now. dir [I]. y] = 'a'; <br/> sort (next. dir, next. dir + 4, CMP); <br/> If (hash1 [Hash (next)]) <br/> continue; <br/> If (hash2 [Hash (next)]) <br/>{< br/> printf ("Yes/N"); <br/> return; <br/>}< br/> hash1 [Hash (next)] = true; <br/> q1.push (next); <br/>}< BR/>}< br/>}</P> <p> while (! Q2.empty () <br/>{< br/> next = now = q2.front (); <br/> q2.pop (); <br/> If (next. step = Step + 1) <br/> break; <br/> for (INT I = 0; I <4; I ++) // four vertices D indicates the pawn, a Indicates blank <br/>{< br/> for (Int J = 0; j <4; j ++) <br/>{< br/> next = now; <br/> X = now. dir [I]. X + dir1 [J] [0]; <br/> Y = now. dir [I]. Y + dir1 [J] [1]; <br/> If (x> = 1 & x <= 8 & Y> = 1 & Y <= 8 & now. map [x] [Y]! = 'D') <br/>{< br/> next. dir [I]. X = x; <br/> next. dir [I]. y = y; <br/> next. step = now. step + 1; <br/> next. map [x] [Y] = 'D'; <br/> next. map [now. dir [I]. x] [now. dir [I]. y] = 'a'; <br/> sort (next. dir, next. dir + 4, CMP); <br/> If (hash2 [Hash (next)]) <br/> continue; <br/> If (hash1 [Hash (next)]) <br/>{< br/> printf ("Yes/N"); <br/> return; <br/>}< br/> hash2 [Hash (next)] = true; <br/> q2.push (next); <br/>}< B R/> for (Int J = 0; j <4; j ++) <br/>{< br/> next = now; <br/> X = now. dir [I]. X + dir2 [J] [0]; <br/> Y = now. dir [I]. Y + dir2 [J] [1]; <br/> If (x> = 1 & x <= 8 & Y> = 1 & Y <= 8 & <br/> now. map [now. dir [I]. X + dir1 [I] [0] [now. dir [I]. Y + dir1 [I] [1] = 'D' <br/> & now. map [x] [Y]! = 'D') <br/>{< br/> next. dir [I]. X = x; <br/> next. dir [I]. y = y; <br/> next. step = now. step + 1; <br/> next. map [x] [Y] = 'D'; <br/> next. map [now. dir [I]. x] [now. dir [I]. y] = 'a'; <br/> sort (next. dir, next. dir + 4, CMP); <br/> If (hash2 [Hash (next)]) <br/> continue; <br/> If (hash1 [Hash (next)]) <br/>{< br/> printf ("Yes/N"); <br/> return; <br/>}< br/> hash2 [Hash (next)] = true; <br/> q2.push (next); <br/>}< br/>} <Br/>}< br/> printf ("No/N"); <br/> return; <br/>}< br/> int main () <br/>{< br/> int X, Y; <br/> while (scanf ("% d", & X, & Y )! = EOF) <br/>{< br/> node SS, ee; <br/> SS. map [x] [Y] = 'D'; <br/> SS. dir [0]. X = x; <br/> SS. dir [0]. y = y; <br/> for (INT I = 1; I <4; I ++) <br/>{< br/> scanf ("% d", & X, & Y); <br/> SS. map [x] [Y] = 'D'; <br/> SS. dir [I]. X = x; <br/> SS. dir [I]. y = y; <br/>}< br/> for (INT I = 0; I <4; I ++) <br/>{< br/> scanf ("% d", & X, & Y); <br/> ee. map [x] [Y] = 'D'; <br/> ee. dir [I]. X = x; <br/> ee. dir [I]. y = y; <br/>}< br/> Start = SS; <br/> end = EE; <br/> sort (start. dir, start. dir + 4, CMP); <br/> sort (end. dir, end. dir + 4, CMP); <br/> BFS (); <br/>}< br/> return 0; <br/>}