Prime Path
Prime Path
POJ 3126
这道题也可以算是到水题吧。搜索加上求素数，这里用的是线性筛。
题目来源: vjudge | POJ
分析
题目要求很简单，给定一个起点和终点，要求用最短的步数到达。每改变一个数算作一步，并且在改变后要求仍然为素数。所以先算出所有素数，然后进行暴力BFS即可。
代码
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| #include <iostream>
#include <queue>
#include <cstring>
using namespace std;
bool prime[10100];
int dist[10100];
int n;
void Prime()
{
truefor (int i = 1; i<=10000; i++)
truetrueprime[i] = true;
trueprime[1] = false;
trueprime[2] = prime[3] = true;
truefor (int i = 2; i<=100; i++)
truetrueif (prime[i])
truetrue{
truetruetrueint j = i*i;
truetruetruewhile (j<=10000)
truetruetrue{
truetruetruetrueprime[j] = false;
truetruetruetruej += i;
truetruetrue}
truetrue}
truereturn;
}
queue<int> q;
int main()
{
truePrime();
truecin >> n;
truefor (int k = 1; k<=n; k++)
true{
truetruememset(dist,0,sizeof(dist));
truetrueint x,y;
truetruecin >> x >> y;
truetrue
truetrueq = queue<int>();
truetruedist[x] = 1;
truetrueq.push(x);
truetruewhile (!q.empty())
truetrue{
truetruetrueint v = q.front(); q.pop();
truetruetrue
truetruetrueif (v==y)
truetruetruetruebreak;
truetruetrueint t = 1;
truetruetruefor (int i = 0; i<4; i++)
truetruetrue{
truetruetruetruefor (int j = 0; j<=9; j++)
truetruetruetrue{
truetruetruetruetrueint z;
truetruetruetruetrueif (t==1)
truetruetruetruetruetruez = v/(t*10)*(t*10) + t*j;
truetruetruetruetrueelse z = v/(t*10)*(t*10) + t*j + v%t;
truetruetruetruetrue
truetruetruetruetrueif (dist[z]==0 && prime[z] && z>=1000)
truetruetruetruetrue{
truetruetruetruetruetruedist[z] = dist[v] + 1;
truetruetruetruetruetrueq.push(z);
truetruetruetruetrue}
truetruetruetrue}
truetruetruetruet *= 10;
truetruetrue}
truetrue}
true
truetruecout << dist[y]-1 << endl;
true}
}
|