Prime Path - POJ 3126 BFS宽度(广度)优先搜索
2014-04-15 18:51
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Prime Path
Description
![](http://poj.org/images/3126_1.jpg)
The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numbers on their offices.
— It is a matter of security to change such things every now and then, to keep the enemy in the dark.
— But look, I have chosen my number 1033 for good reasons. I am the Prime minister, you know!
— I know, so therefore your new number 8179 is also a prime. You will just have to paste four new digits over the four old ones on your office door.
— No, it’s not that simple. Suppose that I change the first digit to an 8, then the number will read 8033 which is not a prime!
— I see, being the prime minister you cannot stand having a non-prime number on your door even for a few seconds.
— Correct! So I must invent a scheme for going from 1033 to 8179 by a path of prime numbers where only one digit is changed from one prime to the next prime.
Now, the minister of finance, who had been eavesdropping, intervened.
— No unnecessary expenditure, please! I happen to know that the price of a digit is one pound.
— Hmm, in that case I need a computer program to minimize the cost. You don't know some very cheap software gurus, do you?
— In fact, I do. You see, there is this programming contest going on... Help the prime minister to find the cheapest prime path between any two given four-digit primes! The first digit must be nonzero, of course. Here is a solution in the case above.
1033
1733
3733
3739
3779
8779
8179
The cost of this solution is 6 pounds. Note that the digit 1 which got pasted over in step 2 can not be reused in the last step – a new 1 must be purchased.
Input
One line with a positive number: the number of test cases (at most 100). Then for each test case, one line with two numbers separated by a blank. Both numbers are four-digit primes (without leading zeros).
Output
One line for each case, either with a number stating the minimal cost or containing the word Impossible.
Sample Input
Sample Output
Source
Northwestern Europe 2006
题意为给你t组数据,然后每组给你a,b两个数,问让你从a改到b需要多少次,要求每次只能改动1个数字,且每次改成的数必须为素数。
本题运用BFS宽度(广度)优先搜索的方法。用一个queue的栈,然后每次从栈中提取出最前面的数,再尝试这个数可以改成什么样的素数,并记录改成这个素数需要多少次(即提出的那个数的改动次数+1),搜索完后删除栈中最前面的那个数。重复上述过程直到搜索到了b数。过程中用数组p[]记录改变到这个素数需要多少次,visited[]用于记录已经搜索过的素数。
代码如下。
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 10483 | Accepted: 5992 |
![](http://poj.org/images/3126_1.jpg)
The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numbers on their offices.
— It is a matter of security to change such things every now and then, to keep the enemy in the dark.
— But look, I have chosen my number 1033 for good reasons. I am the Prime minister, you know!
— I know, so therefore your new number 8179 is also a prime. You will just have to paste four new digits over the four old ones on your office door.
— No, it’s not that simple. Suppose that I change the first digit to an 8, then the number will read 8033 which is not a prime!
— I see, being the prime minister you cannot stand having a non-prime number on your door even for a few seconds.
— Correct! So I must invent a scheme for going from 1033 to 8179 by a path of prime numbers where only one digit is changed from one prime to the next prime.
Now, the minister of finance, who had been eavesdropping, intervened.
— No unnecessary expenditure, please! I happen to know that the price of a digit is one pound.
— Hmm, in that case I need a computer program to minimize the cost. You don't know some very cheap software gurus, do you?
— In fact, I do. You see, there is this programming contest going on... Help the prime minister to find the cheapest prime path between any two given four-digit primes! The first digit must be nonzero, of course. Here is a solution in the case above.
1033
1733
3733
3739
3779
8779
8179
The cost of this solution is 6 pounds. Note that the digit 1 which got pasted over in step 2 can not be reused in the last step – a new 1 must be purchased.
Input
One line with a positive number: the number of test cases (at most 100). Then for each test case, one line with two numbers separated by a blank. Both numbers are four-digit primes (without leading zeros).
Output
One line for each case, either with a number stating the minimal cost or containing the word Impossible.
Sample Input
3 1033 8179 1373 8017 1033 1033
Sample Output
6 7 0
Source
Northwestern Europe 2006
题意为给你t组数据,然后每组给你a,b两个数,问让你从a改到b需要多少次,要求每次只能改动1个数字,且每次改成的数必须为素数。
本题运用BFS宽度(广度)优先搜索的方法。用一个queue的栈,然后每次从栈中提取出最前面的数,再尝试这个数可以改成什么样的素数,并记录改成这个素数需要多少次(即提出的那个数的改动次数+1),搜索完后删除栈中最前面的那个数。重复上述过程直到搜索到了b数。过程中用数组p[]记录改变到这个素数需要多少次,visited[]用于记录已经搜索过的素数。
代码如下。
#include<cstdio> #include<cstring> #include<queue> using namespace std; int f[10010],p[10010],visited[10010]; //f[]用于快速求素数,p[]记录改变到这个素数需要多少次,visited[]用于记录已经搜索过的素数 int BFS(int a,int b) { queue<int> q; q.push(a); p[a]=0; visited[a]=1; while(!q.empty()) { int temp=q.front(); //得到qu最前方的素数 q.pop(); for(int i=0;i<=9;i++) { int y1=(temp/10)*10+i; //个位的改变 if(f[y1]==0 && visited[y1]==0)//判断这个数是否是素数且是否搜索过 { q.push(y1); //是的话将其放入qu的最后 p[y1]=p[temp]+1; //改成这个素数的需要的次数是上一个素数的次数+1 visited[y1]=1; //记录改数搜索已过 } //十位的改变 int y2=(temp/100)*100+temp%10+i*10; if(f[y2]==0 && visited[y2]==0) { q.push(y2); p[y2]=p[temp]+1; visited[y2]=1; } //百位的改变 int y3=(temp/1000)*1000+temp%100+i*100; if(f[y3]==0 && visited[y3]==0) { q.push(y3); p[y3]=p[temp]+1; visited[y3]=1; } //千位的改变 if(i!=0) { int y4=temp%1000+i*1000; if(f[y4]==0 && visited[y4]==0) { q.push(y4); p[y4]=p[temp]+1; visited[y4]=1; } } if(visited[b]) //如果搜素到想要改成的那个数的话,就返回需要的次数 return p[b]; } } return 0; } int main() { int i,j,k,n,a,b; //快速求素数 for(i=2;i<=10010;i++) { if(!f[i]) for(k=2;k*i<10010;k++) f[k*i]=1; } scanf("%d",&n); while(n--) { memset(p,0,sizeof(p)); memset(visited,0,sizeof(visited)); scanf("%d%d",&a,&b); printf("%d\n",BFS(a,b)); } }
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