noip模拟赛 c

分析:一道比较难的爆搜题.首先要把9个块的信息存下来,记录每个块上下左右位置的颜色,然后记录每一排每一列能否操作,之后就是bfs了。在bfs的时候用一个数记录状态,第i位表示原来的第i个块现在在哪个位置,我们可以通过这个状态来解码得到信息,也可以来判重,只是数组开不下,需要用一个map。然后就是如何判断是否连通.首先把所有块还原成一张图,然后我们可以数每个颜色连通的数量,看看是不是等于36就可以了,这里用dfs来判断.
学习了一种比较强的记录状态的方式:记录相对位置,既可以判重,也可以还原图.
#include <cstdio>
#include <map>
#include <queue>
#include <cstring>
#include <iostream>
#include <algorithm> using namespace std; int canx[], cany[], zhuangtai[];
int can[], vis[][], a[][];
map <long long, int> flag; struct node
{
int col[];
}e[]; struct node2
{
long long stu;
int dist;
}; long long hashh(int a, int b, int c, int d, int e, int f, int g, int h, int i)
{
return (((((((a * + b) * + c) * + d) * + e) * + f) * + g) * + h) * + i;
} int dfs(int col, int x, int y, int last)
{
//printf("%d %d %d %d\n", col, x, y, last);
int ret = ;
if (vis[x][y] || (a[x][y] != col && a[x][y] != ) || (a[x][y] == && last == ))
return ret;
vis[x][y] = ;
if (col == a[x][y])
ret++;
if (x > )
ret += dfs(col, x - , y, a[x][y]);
if (x < )
ret += dfs(col, x + , y, a[x][y]);
if (y > )
ret += dfs(col, x, y - , a[x][y]);
if (y < )
ret += dfs(col, x, y + , a[x][y]);
return ret;
} bool check(long long stu)
{
//printf("%lld\n", stu);
int res = ;
memset(can, , sizeof(can));
memset(a, , sizeof(a));
for (int i = ; i >= ; i--)
for (int j = ; j >= ; j--)
{
int x = stu % ;
a[i * ][j * + ] = e[x].col[];
a[i * + ][j * + ] = e[x].col[];
a[i * + ][j * ] = e[x].col[];
a[i * + ][j * + ] = e[x].col[];
stu /= ;
}
for (int i = ; i <= ; i++)
{
for (int j = ; j <= ; j++)
printf("%d ", a[i][j]);
printf("\n");
}
printf("\n");
for (int i = ; i < ; i++)
for (int j = ; j < ; j++)
for (int k = ; k <= ; k++)
if (!can[k] && a[i][j] == k)
{
memset(vis, , sizeof(vis));
res += dfs(k, i, j, );
can[k] = ;
}
if (res == )
return true;
return false;
} void bfs()
{
queue <node2> q;
node2 temp;
temp.stu = ;
temp.dist = ;
q.push(temp);
flag[] = ;
while (!q.empty())
{
node2 u = q.front();
q.pop();
long long stu = u.stu;
int dist = u.dist;
//printf("%lld %d\n", stu, dist);
if (check(stu))
{
printf("%d\n", dist);
return;
}
//printf("flag\n");
for (int i = ; i >= ; i--)
{
zhuangtai[i] = stu % ;
stu /= ;
}
if (!canx[])
{
long long nstu = hashh(zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[]);
//printf("%lld\n", nstu);
if (!flag[nstu])
{
node2 temp;
temp.stu = nstu;
temp.dist = dist + ;
q.push(temp);
}
nstu = hashh(zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[]);
//printf("%lld\n", nstu);
if (!flag[nstu])
{
node2 temp;
temp.stu = nstu;
temp.dist = dist + ;
q.push(temp);
}
}
if (!canx[])
{
long long nstu = hashh(zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[]);
if (!flag[nstu])
{
node2 temp;
temp.stu = nstu;
temp.dist = dist + ;
q.push(temp);
}
nstu = hashh(zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[]);
if (!flag[nstu])
{
node2 temp;
temp.stu = nstu;
temp.dist = dist + ;
q.push(temp);
}
} if (!canx[])
{
long long nstu = hashh(zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[]);
if (!flag[nstu])
{
node2 temp;
temp.stu = nstu;
temp.dist = dist + ;
q.push(temp);
}
nstu = hashh(zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[]);
if (!flag[nstu])
{
node2 temp;
temp.stu = nstu;
temp.dist = dist + ;
q.push(temp);
}
} if (!cany[])
{
long long nstu = hashh(zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[]);
if (!flag[nstu])
{
node2 temp;
temp.stu = nstu;
temp.dist = dist + ;
q.push(temp);
}
nstu = hashh(zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[]);
if (!flag[nstu])
{
node2 temp;
temp.stu = nstu;
temp.dist = dist + ;
q.push(temp);
}
} if (!cany[])
{
long long nstu = hashh(zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[]);
if (!flag[nstu])
{
node2 temp;
temp.stu = nstu;
temp.dist = dist + ;
q.push(temp);
}
nstu = hashh(zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[]);
if (!flag[nstu])
{
node2 temp;
temp.stu = nstu;
temp.dist = dist + ;
q.push(temp);
}
} if (!cany[])
{
long long nstu = hashh(zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[]);
if (!flag[nstu])
{
node2 temp;
temp.stu = nstu;
temp.dist = dist + ;
q.push(temp);
}
nstu = hashh(zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[], zhuangtai[]);
if (!flag[nstu])
{
node2 temp;
temp.stu = nstu;
temp.dist = dist + ;
q.push(temp);
}
}
}
} void init()
{
for (int i = ; i < ; i++)
{
int t;
for (int j = ; j < ; j++)
{
char ch;
cin >> ch;
if (ch == 'R')
e[i].col[j] = ;
if (ch == 'G')
e[i].col[j] = ;
if (ch == 'B')
e[i].col[j] = ;
if (ch == 'O')
e[i].col[j] = ;
}
cin >> t;
if (t == )
{
canx[i / ] = ;
cany[i % ] = ;
}
}
} int main()
{
init();
bfs(); return ;
}
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