题解 SP1841 【PPATH - Prime Path】
模拟赛考到了这个题，但我傻傻的用了\(DFS\)，于是爆了零
后来才想明白，因为搜索树的分支很多，但答案的深度却又没有那么深，所以在这里\(BFS\)，而\(DFS\)一路搜到底的做法则会稳稳地\(T\)飞掉
其他细节请看代码注释
\(code:\)
#include<bits/stdc++.h>
#define maxn 10010
using namespace std;
template<typename T> inline void read(T &x)
{
x=0;char c=getchar();bool flag=false;
while(!isdigit(c)){if(c=='-')flag=true;c=getchar();}
while(isdigit(c)){x=x*10+(c^48);c=getchar();}
if(flag)x=-x;
}
bool 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0,0,0,0,0,0};
bool vis[maxn];//质数表，为搜索判断质数
int m,a,b,ans;
struct node
{
int num;
int cnt;
};
void bfs()
{
queue<node> q;
q.push((node){a,0});
while(!q.empty())
{
int x=q.front().num,cnt=q.front().cnt;
if(x==b)//找到答案，退出循环，因BFS的性质，此时一定为最优解
{
ans=cnt;
break;
}
int base=1,y,to;
for(int i=1;i<=4;++i)//循环将哪一位更改
{
y=x-((x/base)%10)*base;
for(int j=0;j<=9;++j)//循环改成什么数字
{
if(i==4&&j==0) continue;
to=y+j*base;
if(pr[to]&&!vis[to])
{
vis[to]=true;
q.push((node){to,cnt+1});
}
}
base*=10;
}
q.pop();
}
}
void clear()//多组数据，清空
{
ans=-1;
memset(vis,0,sizeof(vis));
}
int main()
{
read(m);
while(m--)
{
clear();
read(a),read(b);
bfs();
if(ans==-1)
puts("Impossible");//若没有答案
else
printf("%d\n",ans);
}
return 0;
}
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