【例题 6-14 UVA-816】Abbott's Revenge

【链接】 我是链接,点我呀:)
【题意】

在这里输入题意

【题解】

预处理出某个方向的左边、前边、右边是哪个方向就好了。 然后就是普通的bfs了。 hash存到某个点,走到这里的方向的最小距离。 dfs输出路径。

【代码】

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

//[where in,point]minimum dis
//0 up 1 down 2 left 3 right
const int dx[4] = { -1,1,0,0 };
const int dy[4] = { 0,0,-1,1 };
const int N = 9;
const int INF = 0x3f3f3f3f;

struct abc {
	int dir, x, y;
};

int bo[N + 5][N + 5][4], xx1, yy1, x2, y2;
int dire[300], newdire[4][300];
int h, t, pre[N*N * 5];
vector <pair<int, pair<int, int> > > v[N + 5][N + 5][4];
vector <pair<int, int> > ans;
abc dl[N*N * 5];

void wujie() {
	puts("");
	puts("  No Solution Possible");
}

bool bfs(int x, int y, int fx) {
	h = 0, t = 1;
	memset(bo, INF, sizeof bo);
	memset(pre, 0, sizeof pre);
	dl[1].dir = fx; dl[1].x = x; dl[1].y = y;
	bo[x][y][fx] = 1;
	if (x==x2 && y == y2) return true;
	while (h < t) {
		h++;
		int tx = dl[h].x, ty = dl[h].y, before = dl[h].dir;
		for (auto temp : v[tx][ty][before]) {
			int xx = tx + temp.second.first, yy = ty + temp.second.second;
			if (xx < 1 || xx > 9 || yy <1 || yy > 9) continue;
			if (bo[xx][yy][temp.first]>bo[tx][ty][before] + 1) {
				bo[xx][yy][temp.first] = bo[tx][ty][before] + 1;
				t++;
				dl[t].x = xx, dl[t].y = yy, dl[t].dir = temp.first;
				pre[t] = h;
				if (xx == x2 && yy == y2) return true;
			}
		}
	}
	return false;
}

void dfs(int now) {
	if (now == 1) {
		ans.push_back(make_pair(xx1, yy1));
		ans.push_back(make_pair(dl[now].x, dl[now].y));
		return;
	}
	dfs(pre[now]);
	ans.push_back(make_pair(dl[now].x, dl[now].y));
}

int main() {
/*	freopen("rush.txt","r",stdin);
	freopen("rush_out.txt","w",stdout);
*/
	dire['N'] = 0, dire['S'] = 1, dire['W'] = 2, dire['E'] = 3;
	newdire[0]['L'] = 2, newdire[0]['F'] = 0, newdire[0]['R'] = 3;
	newdire[1]['L'] = 3, newdire[1]['R'] = 2, newdire[1]['F'] = 1;
	newdire[2]['L'] = 1, newdire[2]['R'] = 0, newdire[2]['F'] = 2;
	newdire[3]['L'] = 0, newdire[3]['R'] = 1, newdire[3]['F'] = 3;
	string T;
	while (cin >> T && T != "END") {
		for (int i = 1; i <= 9; i++)
			for (int j = 1; j <= 9; j++)
				for (int k = 0; k < 4; k++)
					v[i][j][k].clear();
		scanf("%d%d", &xx1, &yy1);
		char s[5];
		scanf("%s", s);
		scanf("%d%d", &x2, &y2);
		int x, y;
		while (scanf("%d", &x) && x) {
			scanf("%d", &y);
			char temp[5];
			while (~scanf("%s", temp) && temp[0] != '*') {
				int from = dire[temp[0]];
				int len = strlen(temp);
				for (int i = 1; i < len; i++) {
					int temp1 = newdire[from][temp[i]];
					v[x][y][from].push_back(make_pair(temp1, make_pair(dx[temp1], dy[temp1])));
				}
			}
		}

		int dir1 = dire[s[0]];
		int tx = xx1 + dx[dir1], ty = yy1 + dy[dir1];
		cout << T;
		if (tx < 1 || tx > 9 || ty < 1 || ty > 9 || !bfs(tx, ty, dir1)) {
			wujie();
		}
		else {
			ans.clear();
			dfs(t);
			bool first;
			for (int i = 0; i < (int)ans.size(); i++) {
				if (i % 10 == 0) {
					puts(""); printf("  ");
					first = true;
				}
				if (!first) putchar(' ');
				first = false;
				printf("(%d,%d)", ans[i].first, ans[i].second);
			}
			puts("");
		}
	}
	return 0;
}