我应该怎么说这道题好呢，只能怪自己英语没有学好吧T^T，为什么造福小众，我决定不翻译题意了，大家自己看吧。。。。

There are several ways to encode an image. In this problem we will consider two representations of an image. We assume that the image consists of black and white pixels. There is at least one black pixel and all black pixels are connected with their sides. Coordinates of black pixels are not less than 1 and not greater than 10. An example of such an image is at the figure.

Both representations describe arrangement of black pixels only.

At the first representation we specify in the first line number of black pixels and coordinates of each black pixel in the following lines. Pixels are listed in order of increasing X. In case of equality of X they are listed in order of increasing Y. Image at the figure is encoded as follows:

6
2 3
2 4
3 3
3 4
4 2
4 3

At the second representation we specify in the first line coordinates of the lowest left black pixel. Each of the following lines contains a description of neighbors for one of the pixels. At first, neighbors of the lowest left pixel are specified, then neighbors of its first neighbor (if it exists) are specified, then neighbors of its second neighbor (if it also exists) follow. When all its neighbors are described the description of the neighbors of its first neighbor follows. The description of the neighbors of its second neighbor follows then and so on.

Each descriptive line contains at most one letter for each neighbor: R for the right, T for the top, L for the left, B for the bottom. If the neighbor was already specified it is not included into the descriptive line and vice-versa. Also there is only one descriptive line for each pixel. Neighbors are listed counter-clockwise starting with the right. Each descriptive line except the last ends with a comma. The last line ends with a full stop. Image at the figure is encoded as follows:

2 3
RT,
RT,
,
B,
,
.

There are no leading or tailing spaces in any representation. There is exactly one space between X and Y coordinates.

Input

One representation of the image will be given to your program in the input.

Output

Your program has to write other representation of the image to the output.

Sample




input
output











6
2 3
2 4
3 3
3 4
4 2
4 3







2 3
RT,
RT,
,
B,
,
.




Problem Source: Third Open USTU Collegiate Programming Contest (PhysTech Cup), March 18, 2000

我只想说重点在最后几句话。。可能只有我没看清楚吧。。。。
看懂题意之后就会发现整个转换的过程其实就是bfs吧，然后就硬上了。一开始我由于没看清楚题意只写了一个转换，wa了几次，后来经过谷哥翻译才发现有这么个坑。
然后读入数据有点麻烦，总之就是我最不擅长的东西，还有要记得是逆时针方向，搜的时候。然后我就因为不明原因一直爆内存。。。。后来的后来，把原本读入字符数组的部分改成读字符串后不知道为什么就过了。。。。白白D了那么久的bug了。。。。

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#include <iostream>
#include <string>
#include <queue>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>

using namespace std;

const int MN = 20;

struct pt{
    int x, y;
};

int n, ans, vx, vy;
bool vis[MN][MN];
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
pt a[MN*MN];
char c[] = {'R', 'T', 'L', 'B'}, b[MN];
queue<pt> q;

bool comp(const pt &pa, const pt &pb) {
    if (pa.x != pb.x) return pa.x < pb.x;
    return pa.y < pb.y;
}

void bfsa() {
    pt now, next;
    string dir;
    while (!q.empty()) {
        now = q.front();
        a[ans] = now;
        cin >> dir;
        for (int i = 0; i < dir.length() - 1; i++) {
            char tc = dir[i];
            int j;
            for (j = 0; j < 4; j++) {
                if (c[j] == tc) break;
            }
            next.x = now.x + dx[j];
            next.y = now.y + dy[j];
            q.push(next);
        }
        q.pop();
        ans++;
    }
}

void bfsb() {
    pt now, next;
    while (!q.empty()) {
        now = q.front();
        for (int i = 0; i < 4; i++) {
            int tx = now.x + dx[i];
            int ty = now.y + dy[i];
            if (vis[tx][ty] == 1) {
                cout << c[i];
                vis[tx][ty] = 0;
                next.x = tx;
                next.y = ty;
                q.push(next);
            }
        }
        ans++;
        if (ans != n)
            cout << ",\n";
        else
            cout << ".\n";
        q.pop();
    }
}

int main() {
    string s;
    getline(cin, s);
    memset(vis, 0, sizeof(vis));
    pt k;
    if (s.find(' ') != string::npos) {
        s.copy(b, s.find(' '), 0);
        vx = atoi(b);
        s.copy(b, 100, s.find(' ') + 1);
        vy = atoi(b);
        pt k;
        k.x = vx;
        k.y = vy;
        q.push(k);
        bfsa();
        sort(a, a+ans, comp);
        cout << ans << endl;
        for (int i = 0; i < ans; i++)
            cout << a[i].x << ' ' << a[i].y << endl;
    } else {
        n = atoi(s.c_str());
        for (int i = 0; i < n; i++) {
            cin >> k.x >> k.y;
            vis[k.x][k.y] = 1;
            if (i == 0) {
                q.push(k);
                cout << k.x << ' ' << k.y << endl;
                vis[k.x][k.y] = 0;
            }
        }
        ans = 0;
        bfsb();
    }
    return 0;
}