Codefoeces 3A. Shortest path of the king

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A. Shortest path of the king
time limit per test
1 second
memory limit per test
64 megabytes
input
standard input
output
standard output

The king is left alone on the chessboard. In spite of this loneliness, he doesn't lose heart, because he has business of national importance. For example, he has to pay an official visit to square t. As the king is not in habit of wasting his time, he wants to get from his current position s to square t in the least number of moves. Help him to do this.

In one move the king can get to the square that has a common side or a common vertex with the square the king is currently in (generally there are 8 different squares he can move to).

Input

The first line contains the chessboard coordinates of square s, the second line — of square t.

Chessboard coordinates consist of two characters, the first one is a lowercase Latin letter (from a to h), the second one is a digit from 1to 8.

Output

In the first line print n — minimum number of the king's moves. Then in n lines print the moves themselves. Each move is described with one of the 8: LRUDLULDRU or RD.

LRUD stand respectively for moves left, right, up and down (according to the picture), and 2-letter combinations stand for diagonal moves. If the answer is not unique, print any of them.

Sample test(s)
input
a8h1
output
7RDRDRDRDRDRDRD

代码:

/*棋盘长度 codeforces 3-A*/#include <stdio.h>#include <algorithm>#include <string>#include <queue>#include <iostream>using namespace std;const int MAXN = 10;struct Grid{    int row;    int col;    Grid(){};    Grid(int x, int y) : row(x), col(y){}    bool operator!=(const Grid &g) const    {        if(row != g.row || col != g.col)            return true;        return false;    }};string Hash[10] = {"LD", "D", "RD", "L", "NONE", "R", "LU", "U", "RU"};int chessBoard[MAXN][MAXN];int dis[MAXN][MAXN];Grid father[MAXN][MAXN];string mov[MAXN][MAXN];bool visit[MAXN][MAXN];void Init(){    for(int i = 0; i < 10; ++i)    {        visit[i][0] = visit[i][9] = visit[9][i] = visit[0][i] = true;    }}void BFS(Grid start){    queue<Grid> que;    que.push(start);    dis[start.row][start.col] = 0;    visit[start.row][start.col] = true;    while(!que.empty())    {        Grid temp = que.front();        int count = 0;        for(int i = -1; i <= 1; ++i)            for(int j = -1; j <= 1; ++j)            {                int x = temp.row + i, y = temp.col + j;                if(!visit[x][y])                {                    visit[x][y] = true;                    mov[x][y] = Hash[count];                    dis[x][y] = dis[temp.row][temp.col] + 1;                    father[x][y] = temp;                    que.push(Grid(x, y));                }                ++count;            }        que.pop();    }}void PrintPath(Grid end, Grid start){    if(end != start)    {        PrintPath(father[end.row][end.col], start);        cout << mov[end.row][end.col] << endl;    }}int main(){#ifdef _LOCAL    freopen("F://input.txt", "r", stdin);#endif    char a, b;    int x, y;    scanf("%c%d", &a, &x);    getchar();    scanf("%c%d", &b, &y);    Grid start(x, a - 'a' + 1), end(y, b - 'a' + 1);    Init();    BFS(start);    printf("%d\n", dis[end.row][end.col]);    PrintPath(end, start);    return 0;}





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