Andrew Stankevich Contest 4 -- Driving Straight, stankevichcontest

Andrew Stankevich Contest 4 -- Driving Straight, stankevichcontest

Question connection

• Question:
  Enter m and n, followed by M-1 rows and 2 * n-1 columns. ''Indicates that it cannot pass, '+' can pass, '-', '|' only allows one direction to pass (in the same direction as the symbol ), calculate the shortest path from the lower left to the upper right corner and output the result: when outputting the result, try to maintain the previous motion direction. At a plus point, the current motion operation is output: R, right turn; L, left turn; F, straight. Output at the beginning: N, go north; E go east
• Analysis:
  The simplest and most short-circuited path is a little effort in output path. Start bfs to find the shortest path in the upper right corner, and start to find the path in the lower left corner.

const int INF = 0x3f3f3f3f;const double EPS = 1e-9;const int MOD = 1000000007;const double PI = acos(-1.0);const int MAXN = 900;char ipt[MAXN][MAXN];int dp[MAXN][MAXN];int n, m;int dx[] = {-1, 0, 1, 0};int dy[] = {0, 1, 0, -1};char judge[] = {'|', '-', '|', '-'};struct Node{    int x, y;    Node(int x = 0, int y = 0) : x(x), y(y) {}} t;bool check(int r, int c, int dir){    return ipt[r][c] == '+' || ipt[r][c] == judge[dir];}void bfs(int sx, int sy){    queue<Node> q;    q.push(Node(sx, sy));    dp[sx][sy] = 0;    while (!q.empty())    {        t = q.front(); q.pop();        REP(i, 4)        {            if (check(t.x, t.y, i))            {                int tx = t.x + dx[i];                int ty = t.y + dy[i];                if (dp[tx][ty] == INF && ipt[tx][ty] != ' ' && ipt[tx][ty] != 0 && check(tx, ty, i))                {                    dp[tx][ty] = dp[t.x][t.y] + 1;                    q.push(Node(tx, ty));                }            }        }    }}void display(int x, int y, int dir){    if (x == 1 && y == m)    {        return;    }    int ldir = (dir + 3) % 4, rdir = (dir + 1) % 4;    int tx, ty;    tx = x + dx[dir]; ty = y + dy[dir];    if (ipt[tx][ty] != 0 && dp[tx][ty] == dp[x][y] - 1)    {        if (ipt[x][y] == '+')            putchar('F');        display(tx, ty, dir);        return;    }    tx = x + dx[ldir]; ty = y + dy[ldir];    if (ipt[tx][ty] != 0 && dp[tx][ty] == dp[x][y] - 1)    {        if (ipt[x][y] == '+')            putchar('L');        display(tx, ty, ldir);        return;    }    tx = x + dx[rdir]; ty = y + dy[rdir];    if (ipt[tx][ty] != 0 && dp[tx][ty] == dp[x][y] - 1)    {        if (ipt[x][y] == '+')            putchar('R');        display(tx, ty, rdir);        return;    }}int main(){    freopen("straight.in", "r", stdin);    freopen("straight.out", "w", stdout);    while (~RII(n, m))    {        n = n * 2 - 1; m = m * 2 - 1;        getchar();        CLR(dp, INF);        CLR(ipt, 0);        FE(i, 1, n)            gets(ipt[i] + 1);        bfs(1, m);        REP(i, 2)        {            int tx = n + dx[i];            int ty = 1 + dy[i];            if (ipt[tx][ty] != 0 && check(n, 1, i) && check(tx, ty, i) && dp[tx][ty] == dp[n][1] - 1)            {                puts(i ? "E" : "N");                display(tx, ty, i);                break;            }        }        puts("");    }    return 0;}

Test data:

2 2+--|||+++2 2+--|+2 2+--+-+--+2 2+++++++++2 2+++-+++++4 4+-+ +-+| |   |+ +-+-+|   -+-+-+-+|     |+-+-+-+