UVa 816 (BFS求最短路)

Posted 2020-08-05 曹孟德

/*816 - Abbott‘s Revenge
---代码完全参考刘汝佳算法入门经典
---strchr() 用来查找某字符在字符串中首次出现的位置，其原型为：char * strchr (const char *str, int c)
---BFS求最短路
--*/
#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;
const int maxn = 10;
struct Node{
	int r, c, dir;
	Node(int a=0, int b=0, int c=0) :r(a), c(b), dir(c){}
};

int dis[maxn][maxn][4];//dis[r][c][dir]保存状态(r,c,dir)到初始状态的距离
int id[256];
Node path[maxn][maxn][4]; //path[r][c][dir]保存了状态(r,c,dir)在BFS树中的父节点
int has[maxn][maxn][4][3];//has[r][c][dir][turn]，表示当前处在(r,c,dir)状态是否可以向turn转弯
int r0, c0, r1, c1, r2, c2, dir;

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

//计算出下一个节点
Node walk(Node&u, int turn){
	//首先计算出下一步的朝向
	int dir = u.dir;
	if (turn == 1)dir = (dir + 3) % 4; //左转，逆时针
	if (turn == 2)dir = (dir + 1) % 4;//右转，顺时针
	return Node(u.r + dr[dir], u.c + dc[dir], dir);
}
bool insid(int r, int c){
	return r >= 1 && c >= 1 && r <= 9 && c <= 9;
}

void print_ans(Node u){
	vector<Node>vec;
	while (dis[u.r][u.c][u.dir] != 0){
		vec.push_back(u);
		u = path[u.r][u.c][u.dir];
	}
	vec.push_back(u);
	vec.push_back(Node(r0, c0, dir));
	int cnt = 0;
	printf("  ");
	for (int i = vec.size() - 1; i >= 0; i--,cnt++){
		if (cnt){
			if (cnt% 10 == 0)printf("\n  ");
			else printf(" ");
		}
		printf("(%d,%d)", vec[i].r, vec[i].c);
	}
	printf("\n");
}

void bfs(){
	queue<Node>Q;
	Node u(r1, c1, dir);
	Q.push(u);
	memset(dis, -1, sizeof(dis));
	dis[r1][c1][dir] = 0;
	while (!Q.empty()){
		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[u.r][u.c][u.dir][i] && insid(v.r, v.c) && dis[v.r][v.c][v.dir] < 0){
				path[v.r][v.c][v.dir] = u;
				dis[v.r][v.c][v.dir] = dis[u.r][u.c][u.dir] + 1;
				Q.push(v);
			}
		}
	}
	printf("  No Solution Possible\n");
}
int main(){
	//01234代表NESW，顺时针方向
	id[‘N‘] = 0;
	id[‘E‘] = 1;
	id[‘S‘] = 2;
	id[‘W‘] = 3;
	//012代表转向
	id[‘F‘] = 0;
	id[‘L‘] = 1;
	id[‘R‘] = 2;
	char s1[21], s2[21];
	while (scanf("%s", s1)&&strcmp(s1,"END")){
		printf("%s\n", s1);
		scanf("%d%d%s%d%d", &r0, &c0, s2, &r2, &c2);
		dir = id[s2[0]];
		r1 = r0 + dr[dir];
		c1 = c0 + dc[dir];
		memset(has, 0, sizeof(has));
		int r, c;
		while (scanf("%d", &r) && r){
			scanf("%d", &c);
			while (scanf("%s", s1) && strcmp(s1, "*")){
				for (int i = 1; i < strlen(s1); i++){
					has[r][c][id[s1[0]]][id[s1[i]]] = 1;
				}
			}
		}
		bfs();
	}
	return 0;
}