uva 816Abbott's Revenge (走迷宫BFS)
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原题链接:
https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=757
参考刘汝佳《算法竞赛入门经典(第二版)》做的(P165)。
耐心看看书上写的步骤挺条理的,也比较好明白。
注意事项:
输出的时候注意空格
代码如下:
#include<iostream>#include<cstring>#include<string>#include<ctype.h>#include<sstream>#include<queue>#include<vector>#include<cstdio>using namespace std;int r0, c0;//入口int r1, c1;//开始走的起始点int r2, c2;//出口int dir;//开始走的方向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[4] = { -1, 0, 1, 0 };const int dc[4] = { 0, 1, 0, -1 };int d[10][10][4];//记录最短路长度,标记面朝dir方向时有没有走过该点bool has_edge[10][10][4][3];//面朝dir方向时有没有这个转向 struct Node{int r, c;int dir;Node (int rr=0,int cc=0,int ddir=0){r = rr;c = cc;dir = ddir;} }p[10][10][4];//记录v的前一个点u,用于找到出口时,将道路逆回去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);}void print_ans( Node u){vector<Node>nodes;while (1){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(" ");//此处一个空格 与下面的空格正好凑成每行前面两个空格printf(" (%d,%d)", nodes[i].r, nodes[i].c);//每个点前面一个空格if (++cnt % 10 == 0)printf("\n");}if (nodes.size() % 10 != 0)printf("\n");}bool inside(int r, int c){return (r > 0 && r <= 9 && c > 0 && 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) //此处意思为: 1.面朝dir方向时有这条边 2.在范围内 3.面朝dir方向没有走过这个点{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");//前面输出两个空格}bool Input(){memset(has_edge, 0, sizeof(has_edge));string name;cin >> name;if (name == "END") return false;//结束条件char ch;cin >> r0>> c0 >> ch >> r2 >> c2;dir = dir_id(ch);r1 = r0 + dr[dir];c1 = c0 + dc[dir];int x, y,dd,tt;while (cin >> x){if (!x) break;cin >> y;string str;while (cin >> str){if (str[0] == '*') break;dd = dir_id(str[0]);for (int i = 1; i < str.length(); i++)has_edge[x][y][dd][turn_id(str[i])] = true;}}cout << name << endl;return true;}int main(){while (Input()){solve();}return 0;}
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