例题6-14 Abbott的复仇（Abbott's Revenge, ACM/ICPC World Finals 2000, UVa816)

原题链接：https://vjudge.net/problem/UVA-816
分类：图
备注：图的最短路(BFS)

前段时间已经做过一些BFS了，做这个就看了看紫书上的翻译，比较关键的是看了下怎么存储路径，大概懂了就很容易了。

代码如下：

#include<string>
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<queue>
#include<map>
using namespace std;
const int dir[4][2] = { {-1,0},{0,1},{1,0},{0,-1} };//NESW
int vised[10][10][4];
int ok[10][10][4][4];
int sr, sc, er, ec;
char toward;
map<char, int>to;
struct node{
	int r, c;
	int t;
	node() :r(0), c(0), t(0) {}
	node(int row, int col, int toward) {
		r = row; c = col; t = toward;
	}
}p[10][10][4];
vector<node>ans;
bool readdir(int row, int col) {
	char ch = getchar();
	if (ch == '*')return false;
	int tt = to[ch]; ch = getchar();
	while (ch != ' ') {
		int tt2 = tt;
		if (ch == 'L')tt2 = (tt + 3) % 4;
		if (ch == 'R')tt2 = (tt + 1) % 4;
		ok[row][col][tt][tt2] = 1;
		ch = getchar();
	}
	return true;
}
bool solve() {
	node head, next;
	head.t = to[toward];
	head.r = sr + dir[to[toward]][0];
	head.c = sc + dir[to[toward]][1];
	queue<node>q;
	q.push(head);
	while (!q.empty()) {
		head = q.front(); q.pop();
		if (head.r == er && head.c == ec) {
			node u = head;
			for (;;) {
				ans.push_back(u);
				if (!p[u.r][u.c][u.t].r)break;
				u = p[u.r][u.c][u.t];
			}
			ans.push_back(node(sr, sc, toward));
			return true;
		}
		for (int i = 0; i < 4; i++) {
			if (!ok[head.r][head.c][head.t][i])continue;
			int rr = head.r + dir[i][0];
			int cc = head.c + dir[i][1];
			if (vised[rr][cc][i])continue;
			next.r = rr; next.c = cc;
			next.t = i;
			vised[rr][cc][i] = 1;
			p[rr][cc][i] = head;
			q.push(next);
		}
	}
	printf("  No Solution Possible\n");
	return false;
}
int main(void) {
	to['N'] = 0; to['E'] = 1;
	to['S'] = 2; to['W'] = 3;
	string name;
	while (cin >> name && name != "END") {
		ans.clear();
		memset(p, 0, sizeof(p));
		memset(vised, 0, sizeof(vised));
		memset(ok, 0, sizeof(ok));
		scanf("%d %d %c %d %d", &sr, &sc, &toward, &er, &ec);
		int row, col;
		while (~scanf("%d", &row) && row) {
			scanf("%d", &col); getchar();
			while (readdir(row, col));
		}
		printf("%s\n", name.c_str());
		if (solve()) {
			for (int i = ans.size() - 1, cnt = 0; i >= 0; i--) {
				if (++cnt % 10 == 1)printf(" ");
				printf(" (%d,%d)", ans[i].r, ans[i].c);
				if (cnt % 10 == 0)printf("\n");
			}
			if (ans.size() % 10 != 0)printf("\n");
		}
	}
	return 0;
}