【Codeforces 242C】King's Path

【链接】 我是链接,点我呀:)

【题意】

让你找到(x0,y0)到(x1,y1)的一条最短路
走过的点必须在所给的n个横向路径上

【题解】

因为n条横向路径上的点最多不会超过10的5次方个,所以我们可以把这10的5次方个点全都
和数字1~10^5一一对应。
然后对于这每一个点,分别于相邻的8个点连边。
然后在无向图上做一个广搜就能找到起点到终点的最短路啦

【代码】

import java.io.*;
import java.util.*;

public class Main {

    static InputReader in;
    static PrintWriter out;

    public static void main(String[] args) throws IOException{
        //InputStream ins = new FileInputStream("E:\\rush.txt");
        InputStream ins = System.in;
        in = new InputReader(ins);
        out = new PrintWriter(System.out);
        //code start from here
        new Task().solve(in, out);
        out.close();
    }

    static int N = (int)1e5;
    static class Task{

        int x0,y0,x1,y1,n;
        int dx[] = {0,0,1,-1,-1,-1,1,1};
        int dy[] = {1,-1,0,0,-1,1,-1,1};
        HashMap<Long,Integer> dic = new HashMap<>();
        int dis[] = new int[N+10];
        ArrayList<Integer> g[] = new ArrayList[N+10];
        Queue<Integer> q = new LinkedList<Integer>();

        int cnt = 0;

        void add(long temp) {
        	if (!dic.containsKey(temp)) {
        		dic.put(temp, ++cnt);
        	}
        }

        long changetohash(int x,int y) {
        	long temp = x;
        	temp = temp << (31);
        	temp = temp + y;
        	return temp;
        }

        public void solve(InputReader in,PrintWriter out) {
        	for (int i = 1;i <= N;i++) g[i] = new ArrayList<>();
        	x0 = in.nextInt();y0 = in.nextInt();x1 = in.nextInt();y1 = in.nextInt();
        	n = in.nextInt();
        	for (int i = 1;i <= n;i++) {
        		int row,cl,cr;
        		row = in.nextInt();
        		cl = in.nextInt();cr = in.nextInt();
        		for (int j = cl;j <= cr;j++) {
        			long temp = changetohash(row,j);
        			add(temp);
        		}
        	}
        	n = cnt;
        	Iterator it = dic.keySet().iterator();
        	while (it.hasNext()) {
        		long value = (long)it.next();
        		long x = value>>>31;
        		long y = 1<<31;y--;
        		y = value&y;
        		int X = dic.get(value);
        		for (int i = 0;i < 8;i++) {
        			int tx = (int)x + dx[i];
        			int ty = (int)y + dy[i];

        			if (dic.containsKey(changetohash(tx, ty))) {
        				int Y = dic.get(changetohash(tx, ty));
        				g[X].add(Y);
        			}
        		}
        	}
        	dis[dic.get(changetohash(x0, y0))] = 1;
        	q.add(dic.get(changetohash(x0, y0)));
        	while (!q.isEmpty()) {
        		int x = q.poll();
        		for (int i = 0;i < (int)g[x].size();i++) {
        			int y = g[x].get(i);
        			if (dis[y]==0) {
        				dis[y] = dis[x]+1;
        				q.add(y);
        			}
        		}
        	}
        	out.println(dis[dic.get(changetohash(x1, y1))]-1);
        }
    }

    static class InputReader{
        public BufferedReader br;
        public StringTokenizer tokenizer;

        public InputReader(InputStream ins) {
            br = new BufferedReader(new InputStreamReader(ins));
            tokenizer = null;
        }

        public String next(){
            while (tokenizer==null || !tokenizer.hasMoreTokens()) {
                try {
                tokenizer = new StringTokenizer(br.readLine());
                }catch(IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }
    }
}