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[5], cany[5], zhuangtai[10];
int can[5], vis[10][10], a[10][10];
map <long long, int> flag;

struct node
{
    int col[5];
}e[10];

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 * 10 + b) * 10 + c) * 10 + d) * 10 + e) * 10 + f) * 10 + g) * 10 + h) * 10 + i;
}

int dfs(int col, int x, int y, int last)
{
    //printf("%d %d %d %d
", col, x, y, last);
    int ret = 0;
    if (vis[x][y] || (a[x][y] != col && a[x][y] != 0) || (a[x][y] == 0 && last == 0))
        return ret;
    vis[x][y] = 1;
    if (col == a[x][y])
        ret++;
    if (x > 0)
        ret += dfs(col, x - 1, y, a[x][y]);
    if (x < 8)
        ret += dfs(col, x + 1, y, a[x][y]);
    if (y > 0)
        ret += dfs(col, x, y - 1, a[x][y]);
    if (y < 8)
        ret += dfs(col, x, y + 1, a[x][y]);
    return ret;
}

bool check(long long stu)
{
    //printf("%lld
", stu);
    int res = 0;
    memset(can, 0, sizeof(can));
    memset(a, 0, sizeof(a));
    for (int i = 2; i >= 0; i--)
        for (int j = 2; j >= 0; j--)
        {
        int x = stu % 10;
        a[i * 3][j * 3 + 1] = e[x].col[0];
        a[i * 3 + 2][j * 3 + 1] = e[x].col[1];
        a[i * 3 + 1][j * 3] = e[x].col[2];
        a[i * 3 + 1][j * 3 + 1] = e[x].col[3];
        stu /= 10;
        }
    for (int i = 0; i <= 8; i++)
    {
        for (int j = 0; j <= 8; j++)
            printf("%d ", a[i][j]);
        printf("
");
    }
    printf("
");
    for (int i = 0; i < 9; i++)
        for (int j = 0; j < 9; j++)
            for (int k = 1; k <= 4; k++)
                if (!can[k] && a[i][j] == k)
                {
                    memset(vis, 0, sizeof(vis));
                    res += dfs(k, i, j, 0);
                    can[k] = 1;
                }
    if (res == 36)
        return true;
    return false;
}

void bfs()
{
    queue <node2> q;
    node2 temp;
    temp.stu = 12345678;
    temp.dist = 0;
    q.push(temp); 
    flag[12345678] = 1;
    while (!q.empty())
    {
        node2 u = q.front();
        q.pop();
        long long stu = u.stu;
        int dist = u.dist;
        //printf("%lld %d
", stu, dist);
        if (check(stu))
        {
            printf("%d
", dist);
            return;
        }
        //printf("flag
");
        for (int i = 8; i >= 0; i--)
        {
            zhuangtai[i] = stu % 10;
            stu /= 10;
        }
        if (!canx[0])
        {
            long long nstu = hashh(zhuangtai[1], zhuangtai[2], zhuangtai[0], zhuangtai[3], zhuangtai[4], zhuangtai[5], zhuangtai[6], zhuangtai[7], zhuangtai[8]);
            //printf("%lld
", nstu);
            if (!flag[nstu])
            {
                node2 temp;
                temp.stu = nstu;
                temp.dist = dist + 1;
                q.push(temp);
            }
            nstu = hashh(zhuangtai[2], zhuangtai[0], zhuangtai[1], zhuangtai[3], zhuangtai[4], zhuangtai[5], zhuangtai[6], zhuangtai[7], zhuangtai[8]);
            //printf("%lld
", nstu);
            if (!flag[nstu])
            {
                node2 temp;
                temp.stu = nstu;
                temp.dist = dist + 1;
                q.push(temp);
            }
        }
        if (!canx[1])
        {
            long long nstu = hashh(zhuangtai[0], zhuangtai[1], zhuangtai[2], zhuangtai[4], zhuangtai[5], zhuangtai[3], zhuangtai[6], zhuangtai[7], zhuangtai[8]);
            if (!flag[nstu])
            {
                node2 temp;
                temp.stu = nstu;
                temp.dist = dist + 1;
                q.push(temp);
            }
            nstu = hashh(zhuangtai[0], zhuangtai[1], zhuangtai[2], zhuangtai[5], zhuangtai[3], zhuangtai[4], zhuangtai[6], zhuangtai[7], zhuangtai[8]);
            if (!flag[nstu])
            {
                node2 temp;
                temp.stu = nstu;
                temp.dist = dist + 1;
                q.push(temp);
            }
        }

        if (!canx[2])
        {
            long long nstu = hashh(zhuangtai[0], zhuangtai[1], zhuangtai[2], zhuangtai[3], zhuangtai[4], zhuangtai[5], zhuangtai[7], zhuangtai[8], zhuangtai[6]);
            if (!flag[nstu])
            {
                node2 temp;
                temp.stu = nstu;
                temp.dist = dist + 1;
                q.push(temp);
            }
            nstu = hashh(zhuangtai[0], zhuangtai[1], zhuangtai[2], zhuangtai[3], zhuangtai[4], zhuangtai[5], zhuangtai[8], zhuangtai[6], zhuangtai[7]);
            if (!flag[nstu])
            {
                node2 temp;
                temp.stu = nstu;
                temp.dist = dist + 1;
                q.push(temp);
            }
        }

        if (!cany[0])
        {
            long long nstu = hashh(zhuangtai[3], zhuangtai[1], zhuangtai[2], zhuangtai[6], zhuangtai[4], zhuangtai[5], zhuangtai[0], zhuangtai[7], zhuangtai[8]);
            if (!flag[nstu])
            {
                node2 temp;
                temp.stu = nstu;
                temp.dist = dist + 1;
                q.push(temp);
            }
            nstu = hashh(zhuangtai[6], zhuangtai[1], zhuangtai[2], zhuangtai[0], zhuangtai[4], zhuangtai[5], zhuangtai[3], zhuangtai[7], zhuangtai[8]);
            if (!flag[nstu])
            {
                node2 temp;
                temp.stu = nstu;
                temp.dist = dist + 1;
                q.push(temp);
            }
        }

        if (!cany[1])
        {
            long long nstu = hashh(zhuangtai[0], zhuangtai[4], zhuangtai[2], zhuangtai[3], zhuangtai[7], zhuangtai[5], zhuangtai[6], zhuangtai[1], zhuangtai[8]);
            if (!flag[nstu])
            {
                node2 temp;
                temp.stu = nstu;
                temp.dist = dist + 1;
                q.push(temp);
            }
            nstu = hashh(zhuangtai[0], zhuangtai[7], zhuangtai[2], zhuangtai[3], zhuangtai[1], zhuangtai[5], zhuangtai[6], zhuangtai[4], zhuangtai[8]);
            if (!flag[nstu])
            {
                node2 temp;
                temp.stu = nstu;
                temp.dist = dist + 1;
                q.push(temp);
            }
        }

        if (!cany[2])
        {
            long long nstu = hashh(zhuangtai[0], zhuangtai[1], zhuangtai[5], zhuangtai[3], zhuangtai[4], zhuangtai[8], zhuangtai[6], zhuangtai[7], zhuangtai[2]);
            if (!flag[nstu])
            {
                node2 temp;
                temp.stu = nstu;
                temp.dist = dist + 1;
                q.push(temp);
            }
            nstu = hashh(zhuangtai[0], zhuangtai[1], zhuangtai[8], zhuangtai[3], zhuangtai[4], zhuangtai[2], zhuangtai[6], zhuangtai[7], zhuangtai[5]);
            if (!flag[nstu])
            {
                node2 temp;
                temp.stu = nstu;
                temp.dist = dist + 1;
                q.push(temp);
            }
        }
    }
}

void init()
{
    for (int i = 0; i < 9; i++)
    {
        int t;
        for (int j = 0; j < 4; j++)
        {
            char ch;
            cin >> ch;
            if (ch == 'R')
                e[i].col[j] = 1;
            if (ch == 'G')
                e[i].col[j] = 2;
            if (ch == 'B')
                e[i].col[j] = 3;
            if (ch == 'O')
                e[i].col[j] = 4;
        }
        cin >> t;
        if (t == 1)
        {
            canx[i / 3] = 1;
            cany[i % 3] = 1;
        }
    }
}

int main()
{
    init();
    bfs();

    return 0;
}