BFS POJ 3126 Prime Path

题目传送门

 /*
题意:从一个数到另外一个数,每次改变一个数字,且每次是素数
BFS:先预处理1000到9999的素数,简单BFS一下。我没输出Impossible都AC,数据有点弱
*/
/************************************************
Author :Running_Time
Created Time :2015-8-2 15:46:57
File Name :POJ_3126.cpp
*************************************************/ #include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std; #define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int MAXN = 1e4 + ;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + ;
bool is_prime[MAXN];
bool vis[MAXN];
int x, y, res;
struct Digit {
int d[];
int step;
}; void solve(void) {
memset (is_prime, true, sizeof (is_prime));
for (int i=; i<=; ++i) {
for (int j=; j<i; ++j) {
if (i % j == ) {
is_prime[i] = false; break;
}
}
}
} int get_num(int *a) {
int ret = ;
for (int i=; i<=; ++i) {
ret = ret * + a[i];
}
return ret;
} void BFS(int u, int v) {
Digit tmp;
for (int i=; i>=; --i) {
tmp.d[i] = u % ; u /= ;
}
tmp.step = ;
queue<Digit> Q; Q.push (tmp); vis[u] = true;
int ans = -;
while (!Q.empty ()) {
Digit x = Q.front (); Q.pop ();
int m = get_num (x.d);
if (m == v) {
ans = x.step; break;
}
for (int i=; i<=; ++i) {
for (int j=; j<=; ++j) {
if (i == && j == ) continue;
if (x.d[i] != j) {
Digit y = x;
y.d[i] = j; m = get_num (y.d);
if (is_prime[m] && !vis[m]) {
vis[m] = true; y.step++; Q.push (y);
}
}
}
}
}
if (ans == -) puts ("Impossible");
else printf ("%d\n", ans);
} int main(void) { //POJ 3126 Prime Path
solve ();
int T; scanf ("%d", &T);
while (T--) {
scanf ("%d%d", &x, &y);
memset (vis, false, sizeof (vis));
BFS (x, y);
} return ;
}
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