poj 3126 （素数筛）

最新推荐文章于 2020-12-09 10:16:52 发布

FawkesLi

最新推荐文章于 2020-12-09 10:16:52 发布

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分类专栏：算法-数论算法文章标签：算法

本文链接：https://blog.youkuaiyun.com/Fawkess/article/details/102884140

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本文介绍了一种使用BFS算法寻找两个素数之间的最短路径的方法，并通过预处理生成了一个素数列表来加速判断过程。代码中包含了初始化素数列表、BFS算法实现以及主函数调用。

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#include <iostream>
#include <cstdio>
#include <iomanip>
#include <string>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <set>
#include <vector>
#include <map>
#include <algorithm>
#include <cmath>
#include <stack>
#include <stdlib.h>
#include <stdio.h>

#define INF 0x3f3f3f3f
#define LINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define ull unsigned long long
#define uint unsigned int
#define l(x) x<<1
#define r(x) x<<1|1
#define ms(a,b) memset(a,b,sizeof(a))
//#define ms(a,b,c,d) memset(a,b,c*sizeof(d))

using namespace std;

int prime[11111];
void init() {
	for (int i = 3; i < 100; i+=2) {
		if (prime[i] == 1) continue;
		for (int j = 3; j < 10000; j+=2) {
			if (i*j > 10000) break;
			prime[i*j] = 1;
		}
	}
}

struct node {
	int x;
	int times;
};

int t, n1, n2,vis[11111];

int bfs() {
	ms(vis, 0);
	queue<node> mq;
	node now, temp;
	now.x = n1;
	now.times = 0;
	mq.push(now);
	int tn;

	while (!mq.empty()) {
		now = mq.front();
		mq.pop();
		if (now.x == n2) return now.times;

		for (int i = 1; i < 10; i += 2) {
			tn = i - (now.x % 10);
			if (tn == 0) continue;
			temp.x = now.x + tn;
			if (prime[temp.x] == 0 && vis[temp.x] == 0) {
				vis[temp.x] = 1;
				temp.times = now.times + 1;
				mq.push(temp);
			}
		}

		for (int i = 0; i < 10; i++) {
			tn = i - (now.x % 100) / 10 ;
			if (tn == 0) continue;
			temp.x = now.x + tn * 10;
			if (prime[temp.x] == 0 && vis[temp.x] == 0) {
				vis[temp.x] = 1;
				temp.times = now.times + 1;
				mq.push(temp);
			}
		}

		for (int i = 0; i < 10; i++) {
			tn = i - (now.x % 1000) / 100;
			if (tn == 0) continue;
			temp.x = now.x + tn * 100;
			if (prime[temp.x] == 0 && vis[temp.x] == 0) {
				vis[temp.x] = 1;
				temp.times = now.times + 1;
				mq.push(temp);
			}
		}

		for (int i = 1; i < 10; i++) {
			tn = i - now.x/ 1000;
			if (tn == 0) continue;
			temp.x = now.x + tn * 1000;
			if (prime[temp.x] == 0 && vis[temp.x] == 0) {
				vis[temp.x] = 1;
				temp.times = now.times + 1;
				mq.push(temp);
			}
		}
	}
	return -1;
}

int main() {
	ms(prime, 0);
	init();
	cin >> t;
	while (t--) {
		cin >> n1 >> n2;
		cout << bfs() << endl;
	}

	return 0;
}