UVA 816 Abbott's Revenge(带方向的bfs）

​​UVA 816​​

题意

给你一幅图， 求起点到终点最短路，图中每个节点进入的朝向跟出去的方向题目会给出。

分析

用一个三元组来保存状态(r, c, dir),代表当前位置在r, c。面朝dir方向。

#include<bits/stdc++.h>
using namespace std;

struct Node
{
    int r, c, dir; // 站在(r,c)，面朝方向dir(0~3分别表示N, E, S, W)
    Node(int r = 0, int c = 0, int dir = 0) : r(r), c(c), dir(dir) {}
};

const int maxn = 10;
const char *dirs = "NESW"; // 顺时针旋转
const char *turns = "FLR";

int has_edge[maxn][maxn][4][3];
int d[maxn][maxn][4];
Node p[maxn][maxn][4];
int r0, c0, dir, r1, c1, r2, c2;

int dir_id(char c) { return strchr(dirs, c) - dirs; }
int turn_id(char c) { return strchr(turns, c) - turns; }

const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};

Node walk(const Node &u, int turn)
{
    int dir = u.dir;
    if (turn == 1)
        dir = (dir + 3) % 4; // 逆时针
    if (turn == 2)
        dir = (dir + 1) % 4; // 顺时针
    return Node(u.r + dr[dir], u.c + dc[dir], dir);
}

bool inside(int r, int c)
{
    return r >= 1 && r <= 9 && c >= 1 && c <= 9;
}

bool read_case()
{
    char s[99], s2[99];
    if (scanf("%s%d%d%s%d%d", s, &r0, &c0, s2, &r2, &c2) != 6)
        return false;
    printf("%s\n", s);

    dir = dir_id(s2[0]);
    r1 = r0 + dr[dir];
    c1 = c0 + dc[dir];

    memset(has_edge, 0, sizeof(has_edge));
    for (;;)
    {
        int r, c;
        scanf("%d", &r);
        if (r == 0)
            break;
        scanf("%d", &c);
        while (scanf("%s", s) == 1 && s[0] != '*')
        {
            for (int i = 1; i < strlen(s); i++)
                has_edge[r][c][dir_id(s[0])][turn_id(s[i])] = 1;
        }
    }
    return true;
}

void print_ans(Node u)
{
    // 从目标结点逆序追溯到初始结点
    vector<Node> nodes;
    for (;;)
    {
        nodes.push_back(u);
        if (d[u.r][u.c][u.dir] == 0)
            break;
        u = p[u.r][u.c][u.dir];
    }
    nodes.push_back(Node(r0, c0, dir));

    // 打印解，每行10个
    int cnt = 0;
    for (int i = nodes.size() - 1; i >= 0; i--)
    {
        if (cnt % 10 == 0)
            printf(" ");
        printf(" (%d,%d)", nodes[i].r, nodes[i].c);
        if (++cnt % 10 == 0)
            printf("\n");
    }
    if (nodes.size() % 10 != 0)
        printf("\n");
}

void solve()
{
    queue<Node> q;
    memset(d, -1, sizeof(d));
    Node u(r1, c1, dir);
    d[u.r][u.c][u.dir] = 0;
    q.push(u);
    while (!q.empty())
    {
        Node u = q.front();
        q.pop();
        if (u.r == r2 && u.c == c2)
        {
            print_ans(u);
            return;
        }
        for (int i = 0; i < 3; i++)
        {
            Node v = walk(u, i);
            if (has_edge[u.r][u.c][u.dir][i] && inside(v.r, v.c) && d[v.r][v.c][v.dir] < 0)
            {
                d[v.r][v.c][v.dir] = d[u.r][u.c][u.dir] + 1;
                p[v.r][v.c][v.dir] = u;
                q.push(v);
            }
        }
    }
    printf("  No Solution Possible\n");
}

int main()
{
    while (read_case())
    {
        solve();
    }
    return 0;
}