poj_3126

源链接：http://poj.org/problem?id=3126

题目大意就是给你两个素数a,b，问从a开始，只改变其中的一个数字，使它变为另一个素数，一直如此，问是否能够变为b，如果能，输出最小的步数，否则输出impossible

bfs即可。从a开始，枚举改变每一个位置上的数，是素数的话就放入队列，直到找到b，要注意，千位上的数字枚举时得从1开始。

#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<iomanip>#include<vector>#include<time.h>#include<queue>#include<stack>#include<iterator>#include<math.h>#include<stdlib.h>#include<limits.h>#include<map>//#define ONLINE_JUDGE#define eps 1e-8#define INF 0x7fffffff#define FOR(i,a) for((i)=0;i<(a);(i)++)#define MEM(a) (memset((a),0,sizeof(a)))#define sfs(a) scanf("%s",a)#define sf(a) scanf("%d",&a)#define sfI(a) scanf("%I64d",&a)#define pf(a) printf("%d\n",a)#define pfI(a) printf("%I64d\n",a)#define pfs(a) printf("%s\n",a)#define sfd(a,b) scanf("%d%d",&a,&b)#define sft(a,b,c)scanf("%d%d%d",&a,&b,&c)#define for1(i,a,b) for(int i=(a);i<b;i++)#define for2(i,a,b) for(int i=(a);i<=b;i++)#define for3(i,a,b)for(int i=(b);i>=a;i--)#define MEM1(a) memset(a,0,sizeof(a))#define MEM2(a) memset(a,-1,sizeof(a))const double PI=acos(-1.0);template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}template<class T> inline T Min(T a,T b){return a<b?a:b;}template<class T> inline T Max(T a,T b){return a>b?a:b;}using namespace std;#define ll __int64int n,m;int L,R,C;#define Mod 1000000007#define N 110#define M 1000100int dirx[4] = {0,0,1,-1};int diry[4] = {1,-1,0,0};int a,b;int prime[10010];int vis[10010];void Init(){memset(prime,0,sizeof prime);prime[2] = 0;for(int i=2;i*i<10010;i++){if(!prime[i]){for(int j=i;i*j<10010;j++)prime[i*j] = 1;}}}struct Node{int val,step;};int bfs(){queue<Node> q;Node cur,next;cur.val = a;cur.step = 0;q.push(cur);memset(vis,0,sizeof vis);vis[a] = 1;while(!q.empty()){cur = q.front();int tmp = cur.val;q.pop();int d0 = tmp%10;int d1 = (tmp/10)%10;int d2 = (tmp/100)%10;int d3 = (tmp/1000)%10;for(int i=0;i<10;i++){if(i == d0) continue; next.val = d3*1000+d2*100+d1*10+i; next.step = cur.step + 1; if(next.val == b) return next.step;if(!prime[next.val]&&!vis[next.val]){vis[next.val] = 1;q.push(next);}}for(int i=0;i<10;i++){if(i == d1) continue;next.val = d3*1000+d2*100+i*10+d0;next.step = cur.step+1;if(next.val == b)return next.step;if(!prime[next.val] && !vis[next.val]){vis[next.val] = 1;q.push(next);}}for(int i=0;i<10;i++){if(i == d2) continue;next.val = d3*1000+i*100+d1*10+d0;next.step = cur.step+1;if(next.val == b)return next.step;if(!prime[next.val] && !vis[next.val]){vis[next.val] = 1;q.push(next);}}for(int i=1;i<10;i++){if(i == d3) continue;next.val = i*1000+d2*100+d1*10+d0;next.step = cur.step+1;if(next.val == b)return next.step;if(!prime[next.val] && !vis[next.val]){vis[next.val] = 1;q.push(next);}}}return 0;}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);//  freopen("out.txt","w",stdout);#endif    int t;    sf(t);    Init();    while(t--){    sfd(a,b);    if(a == b){    printf("0\n");    continue;    }    int ans = bfs();    if(ans)    printf("%d\n",ans);    else    printf("Impossible\n");    }return 0;}