NYOJ 题目592 spiral grid (广搜+素数预处理+大表)

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

题意：给出一张100*100的蛇形地图，给出两个数据，问从第一个数据走到第二个数据需要多少步，不能走到素数格子，所以终点是素数的直接impossible#include<cstdio>
#include<iostream>
#include<queue>
#include<cstring>
using namespace std;
struct node{
int x;
int y;
int step;
};
queue <struct node> q;
int map[105][105];
int vis[105][105];
int prime_0[10005];//素数是0
int go[4][2]={1,0,0,1,-1,0,0,-1};
int n=100;
void initial(){//初始化，数组清空，队列清空
memset(vis,0,sizeof(vis));
while(!q.empty()){
q.pop();
}
}
void find_prime(){//预处理，把素数先全都找到
prime_0[1]=1;
int i,j;
for(i=2;i<=10005;i++){
if(prime_0[i]==0)
for(j=i+i;j<=10005;j+=i)
prime_0[j]=1;
}
}
struct node locate(int aa){//找到一个数字的坐标
int i,j;
struct node bb;
bb.step=0;
for(i=1;i<=100;i++)
for(j=1;j<=100;j++){
if(map[i][j]==aa){
bb.x=i;
bb.y=j;
break;
}
}
return bb;
}
void build_map(int maxn){//地图打表
int i,j;
for(i=1;i<=n/2;i++){
for(j=1;j<=n-2*i+2;j++)
map[i][j+i-1]=maxn--;
for(j=1;j<=n-2*i;j++)
map[i+j][n-i+1]=maxn--;
for(j=n-i+1;j>=i;j--)
map[n-i+1][j]=maxn--;
for(j=n-i;j>=i+1;j--)
map[j][i]=maxn--;
}
}

int bfs(int x1,int y1,int en){//广度搜索
int i;

struct node e={x1,y1,0};
q.push(e);
vis[x1][y1]=1;
while(!q.empty()){
e=q.front();
if(map[e.x][e.y]==en)break;
q.pop();
for(i=0;i<4;i++){
int xx=e.x+go[i][0];
int yy=e.y+go[i][1];
if(xx>=1&&xx<=100&&yy>=1&&yy<=100&&vis[xx][yy]==0&&prime_0[map[xx][yy]]){
struct node ee={xx,yy,e.step+1};
q.push(ee);
vis[xx][yy]=1;
}
}
}
if(q.empty())return -1;
else return e.step;
}
int main(){
int case_num=1;
int i,j,num,st,en;
build_map(n*n);
find_prime();
while(scanf("%d%d",&st,&en)!=EOF){
if(!prime_0[en]){
printf("Case %d: impossible\n",case_num++);
continue;
}//终点是素数，直接impossible
struct node start=locate(st);//找到起点坐标
initial();//初始化
int ans=bfs(start.x,start.y,en);
if(ans==-1)printf("Case %d: impossible\n",case_num++);
else printf("Case %d: %d\n",case_num++,ans);
}
return 0;
}


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