hdu 1401 zoj1505 pku 1198 双向广搜

]/*
对于双广一直很害怕，一直不敢去写。。。
对于状态的保存总是感觉很陌生，可是BFS DFS 中的状态很重要，好好学状态
以前就想写这个题目。。。
终于AC了。。
*/
#include <iostream>
#include <cstdio>
#include <bitset>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int N =( (1 << 24) + 10);
bitset<N> hash1; // 说这个能节省内存
bitset<N> hash2;
int dir1[4][2] = { {-1, 0}, {1, 0}, {0, 1}, {0, -1} };
int dir2[4][2] = { {-2, 0}, {2, 0}, {0, 2}, {0, -2} };
char map1[9][9], map2[9][9];
struct Dir
{
int x, y;
};
struct Node
{
Dir dir[4];
int step;
char map[9][9];
Node()
{
for(int i = 0; i < 9; i++)
for(int j = 0; j < 9; j++)
map[i][j] = 'A';
step = 0;
}
};
Node start, end;
inline bool cmp(const Dir &a, const Dir&b)
{//对坐标排序避免不必要的入队列
if(a.x == b.x)
return a.y < b.y;
else
return a.x < b.x;
}
inline int  Hash(const Node &node)
{//将坐标转化为一维的
int num;
int sum = 0;
for(int i = 0; i < 4; i++)
{
num = (node.dir[i].x - 1) * 8 + node.dir[i].y;
sum = 64 * sum + num ;
}
return sum;
}
void BFS()
{
queue<Node> Q1, Q2;
Q1.push(start);
Q2.push(end);
Node now, next;
hash1.reset();
hash2.reset();
hash1[ Hash(start)] = true;
hash2[ Hash(end)] = true;
int x, y;
for(int step = 0; step < 4; step++)
{
while(!Q1.empty())
{
next = now = Q1.front();
Q1.pop();
if(next.step > 4)
break;
for(int i = 0; i < 4; i++) // 四个点 D表示棋子，A表示空白
{
for(int j = 0; j < 4; j++)
{// 先搜一步的方向
next = now;
x = now.dir[i].x + dir1[j][0];
y = now.dir[i].y + dir1[j][1];
if(x >= 1 && x <= 8 && y >= 1 && y <= 8 && now.map[x][y] != 'D')
{
next.dir[i].x = x;
next.dir[i].y = y;
next.step = now.step + 1;
next.map[x][y] = 'D';
next.map[now.dir[i].x][now.dir[i].y] = 'A';
sort(next.dir, next.dir + 4 , cmp);
if(hash1[ Hash(next) ])
continue;
if(hash2[ Hash(next) ] )
{//如果该状态在end 搜过来hash过的就说明存在
printf("YES/n");
return ;
}
hash1[ Hash(next) ] = true;
Q1.push(next);
}
}
for(int j = 0; j < 4; j++)
{
next = now;
x = now.dir[i].x + dir2[j][0];
y = now.dir[i].y + dir2[j][1];
if(x >= 1 && x <= 8 && y >= 1 && y <= 8 &&
now.map[now.dir[i].x + dir1[i][0] ][now.dir[i].y + dir1[i][1] ] == 'D'
&& now.map[x][y] != 'D')
{
next.dir[i].x = x;
next.dir[i].y = y;
next.step = now.step + 1;
next.map[x][y] = 'D';
next.map[now.dir[i].x][now.dir[i].y] = 'A';
sort(next.dir, next.dir + 4 , cmp);
if(hash1[ Hash(next) ])
continue;
if(hash2[ Hash(next) ] )
{
printf("YES/n");
return ;
}
hash1[ Hash(next) ] = true;
Q1.push(next);
}
}
}
}

while(!Q2.empty())
{
next = now = Q2.front();
Q2.pop();
if(next.step == step + 1)
break;
for(int i = 0; i < 4; i++) // 四个点 D表示棋子，A表示空白
{
for(int j = 0; j < 4; j++)
{
next = now;
x = now.dir[i].x + dir1[j][0];
y = now.dir[i].y + dir1[j][1];
if(x >= 1 && x <= 8 && y >= 1 && y <= 8 && now.map[x][y] != 'D')
{
next.dir[i].x = x;
next.dir[i].y = y;
next.step = now.step + 1;
next.map[x][y] = 'D';
next.map[now.dir[i].x][now.dir[i].y] = 'A';
sort(next.dir, next.dir + 4 , cmp);
if(hash2[ Hash(next)])
continue;
if(hash1[ Hash(next) ] )
{
printf("YES/n");
return ;
}
hash2[ Hash(next) ] = true;
Q2.push(next);
}
}
for(int j = 0; j < 4; j++)
{
next = now;
x = now.dir[i].x + dir2[j][0];
y = now.dir[i].y + dir2[j][1];
if(x >= 1 && x <= 8 && y >= 1 && y <= 8 &&
now.map[now.dir[i].x + dir1[i][0] ][now.dir[i].y + dir1[i][1] ] == 'D'
&& now.map[x][y] != 'D')
{
next.dir[i].x = x;
next.dir[i].y = y;
next.step = now.step + 1;
next.map[x][y] = 'D';
next.map[now.dir[i].x ][now.dir[i].y ] = 'A';
sort(next.dir, next.dir + 4 , cmp);
if(hash2[ Hash(next)])
continue;
if(hash1[ Hash(next) ] )
{
printf("YES/n");
return ;
}
hash2[ Hash(next) ] = true;
Q2.push(next);
}
}
}
}
}
printf("NO/n");
return ;
}
int main()
{
int x, y;
while(scanf("%d %d", &x, &y) != EOF)
{
Node ss, ee;
ss.map[x][y] = 'D';
ss.dir[0].x = x;
ss.dir[0].y = y;
for(int i = 1; i < 4; i++)
{
scanf("%d %d", &x, &y);
ss.map[x][y] = 'D';
ss.dir[i].x = x;
ss.dir[i].y = y;
}
for(int i = 0; i < 4; i++)
{
scanf("%d %d", &x, &y);
ee.map[x][y] = 'D';
ee.dir[i].x = x;
ee.dir[i].y = y;
}
start = ss;
end = ee;
sort(start.dir, start.dir + 4 , cmp);
sort(end.dir, end.dir + 4, cmp);
BFS();
}
return 0;
}