NYOJ592 spiral grid 【BFS】

spiral grid

时间限制： 2000 ms  |  内存限制： 65535 KB

难度： 4

描述

Xiaod has recently discovered the grid named "spiral grid".
Construct the grid like the following figure. (The grid is actually infinite. The figure is only a small part of it.)


Considering traveling in it, you are free to any cell containing a composite number or 1, but traveling to any cell containing a prime number is disallowed. In addition, traveling from a prime number is disallowed, either. You can travel up, down, left or right, but not diagonally. Write a program to find the length of the shortest path between pairs of nonprime numbers, or report it's impossible.

输入

Each test case is described by a line of input containing two nonprime integer 1 <=x, y<=10,000.

输出

For each test case, display its case number followed by the length of the shortest path or "impossible" (without quotes) in one line.

样例输入

1 4
9 32
10 12

样例输出

Case 1: 1
Case 2: 7
Case 3: impossible

构图的时候要格外细心。

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <queue>
#define abs(a) (a) > 0 ? (a) : -(a)
using std::queue;

int map[102][102], vis[102][102], prime[10002] = {1, 1};
int x, y, mov[][2] = {0, 1, 0, -1, 1, 0, -1, 0};
struct Node{
	int x, y, steps;
};
queue<Node> Q;

void preprocess(){
	//画图的时候要细心
	int count = 10000, x = 2, y = 2;
	for(int i = 1; i <= 100; ++i)
		map[1][i] = count--;
	for(int i = 2; i <= 100; ++i)
		map[i][100] = count--;
	for(int i = 99; i > 0; --i)
		map[100][i] = count--;
	for(int i = 99; i > 1; --i)
		map[i][1] = count--;
	map[2][2] = count--;
	while(count > 0){
		while(!map[x][y+1]) map[x][++y] = count--;
		while(!map[x+1][y]) map[++x][y] = count--;
		while(!map[x][y-1]) map[x][--y] = count--;
		while(!map[x-1][y]) map[--x][y] = count--;
	}	
}

Node getX(){
	Node t = {0};
	for(int i = 1; i < 101; ++i)
		for(int j = 1; j < 101; ++j){
			if(map[i][j] == x){
				t.x = i; t.y = j;
				return t;
			}
		}
}

void getPrime(){
	for(int i = 2; i <= 100; ++i){
		if(prime[i]) continue;
		for(int j = i * i; j <= 10000; j += i)
			prime[j] = 1;
	}
}

int check(Node t){
	if(t.x < 1 || t.y < 1 || t.x > 100 || t.y > 100)
		return 0;
	if(vis[t.x][t.y] || !prime[map[t.x][t.y]])
		return 0;
	return 1;
}

void BFS(){
	Node temp, now;
	while(!Q.empty()) Q.pop();
	now = getX();
	if(x == y){
		printf("0\n"); return;
	}
	Q.push(now);
	
	while(!Q.empty()){
		now = Q.front();
		Q.pop();
		
		for(int i = 0; i < 4; ++i){
			temp = now;
			temp.x += mov[i][0];
			temp.y += mov[i][1];
			++temp.steps;
			
			if(check(temp)){
				if(map[temp.x][temp.y] == y){
					printf("%d\n", temp.steps);
					return;
				}
				vis[temp.x][temp.y] = 1;
				Q.push(temp);
			}
		}
	}
	printf("impossible\n");
}

int main(){
	getPrime();
	preprocess();
	int times = 1;
	while(scanf("%d%d", &x, &y) == 2){
		printf("Case %d: ", times++);
		memset(vis, 0, sizeof(vis));
		BFS();
	}
	return 0;
}