Abbott的复仇 Uva816

这道题无疑是学bfs的一道经典题，难度不是很大，就是要注意一些细节~
不小心把一下的NESW打成NEWS真是作死，，

const char* dirs = “NESW”;

下面贴码~

#include <iostream>
#include <cstring>
#include <queue>
#include <cstdio>
using namespace std;

const char* dirs = "NESW";
const char* turns = "FLR";
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};

struct Node{
    int r;
    int c;
    int dir;
    Node(int r=0, int c=0, int dir=0){
        this->c = c;
        this->r = r;
        this->dir = dir;
    }
};

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

int r0, c0, dir, r1, c1, r2,c2;//起点, 起点朝向, 终点.
const int maxn = 20;
int d[maxn][maxn][maxn];
Node p[maxn][maxn][maxn];
int has_edge[maxn][maxn][maxn][maxn];

void print_ans(Node u){
    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));

    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");
}

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

void solve(){
    queue<Node> q;
    memset(d, -1, sizeof(d));
    Node u(r1, c1, dir);
    d[u.r][u.c][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()
{
    memset(p, 0, sizeof(p));
    memset(has_edge, 0, sizeof(has_edge));  
    char c;
    scanf("%d%d", &r0, &c0);
    cin >> c;
    scanf("%d%d%",&r2, &c2);
    dir = dir_id(c);
    r1 = r0 + dr[dir]; c1 = c0 + dc[dir];
    //构建整个图
    int row, col;
    while(cin >> row){
        if(row == 0) break;
        cin >> col;
        string s;
        while(cin >> s){
            if(s == "*") break;
            int dirid = dir_id(s[0]);
            for(int i = 1; i <s.length(); i++){
                has_edge[row][col][dirid][turn_id(s[i])] = 1;
            }
        }
    }
    solve();
    return 0;
}