百练OJ-Prime Path-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_34392405/article/details/79744769

本文介绍了一种基于广度优先搜索（BFS）的算法，用于寻找两个四位素数之间的最短转换路径。该算法首先生成所有小于100的素数列表，然后通过改变每个数位来生成新的素数并进行遍历，直到找到目标素数。

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#include<queue>
#include<iostream>
#include<vector>
#include<math.h>
#include<cstring>
using namespace std;

vector<int>primes;

void makeprimes() {
	primes.push_back(2);
	for (int i = 3; i < 100; i += 2) {
		bool fg = true;
		for (int j = 0; j<primes.size()&&primes[j] <= sqrt(i); j++) {
			if (i%primes[j] == 0) {
				fg = false;
				break;
			}
		}
		if (fg) {
			primes.push_back(i);
		}
	}
}

bool isPrime(int n) {
	for (int i = 0; i < primes.size() && primes[i] <= sqrt(n); i++) {
		if (n%primes[i] == 0) {
			return 0;
		}
	}
	return 1;
}

void BFS(int a,int b) {
	queue<int>q;
	int cost[10000];
	memset(cost, -1, sizeof(cost));

	q.push(a);
	cost[a] = 0;
	while (!q.empty()) {
		int now = q.front();
		q.pop();
		int gewei = now % 10;
		int shiwei = int(now / 10)%10;
		int baiwei = int(now / 100) % 10;
		int qianwei = int(now / 1000);

		int costnow = cost[now];

		if (now == b) {
			cout<<costnow<<endl;
			return;
		}
		for (int i = 1; i < 10; i++) {//千位
			if (i == qianwei)
				continue;
			int temp = i * 1000 + baiwei * 100 + shiwei * 10 + gewei;
			if (cost[temp] == -1 && isPrime(temp)) {
				cost[temp] = costnow + 1;
				q.push(temp);
			}
		}
		for (int i = 0; i < 10; i++) {//百位
			if (i == baiwei)
				continue;
			int temp = qianwei * 1000 + i * 100 + shiwei * 10 + gewei;
			if (cost[temp] == -1 && isPrime(temp)) {
				cost[temp] = costnow + 1;
				q.push(temp);
			}
		}
		for (int i = 0; i < 10; i++) {//十位
			if (i == shiwei) {
				continue;
			}
			int temp = qianwei * 1000 + baiwei * 100 + i * 10 + gewei;
			if (cost[temp] == -1 && isPrime(temp)) {
				cost[temp] = costnow + 1;
				q.push(temp);
			}
		}
		for (int i = 1; i < 10; i += 2) {//个位
			if (i == gewei) { continue; }
			int temp = qianwei * 1000 + baiwei * 100 + shiwei * 10 + i;
			if (cost[temp] == -1 && isPrime(temp)) {
				cost[temp] = costnow + 1;
				q.push(temp);
			}
		}
	}
}

int main()
{
	int a, b;
	int n;
	cin >> n;
	makeprimes();

	while (n--) {
		cin >> a >> b;
		BFS(a, b);
	}
    return 0;
}