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  • POJ 3126 Prim Parh

    Prime Path
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 13905   Accepted: 7841

    Description

    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


    这个题一看知道初始状态,知道末状态,准备用双向BFS来做,但是老是WA,也一直苦于找不到WA数据。所以暂时先用普通的单向BFS来做,回头再用双向的做一做。

    这个题只需要把可以达到的质数入队就可以了。注意判重。

    /*************************************************************************
    	> File Name: Prim_Path_by_bfs.cpp
    	> Author: Zhanghaoran0
    	> Mail: chiluamnxi@gmail.com
    	> Created Time: 2015年07月29日 星期三 08时37分18秒
     ************************************************************************/
    
    #include <iostream>
    #include <algorithm>
    #include <cstdio>
    #include <cstring>
    #include <math.h>
    using namespace std;
    
    struct node{
        int x;
        int step;
    }q[10000];
    bool flag[10000];
    int T;
    int head, tail;
    int start, end;
    bool judge_prim(int x){
        if(x == 1 || x == 0)
            return false;
        if(x == 2)
            return true;
        for(int i = 2; i <= (int)sqrt(x); i ++){
            if(x % i == 0)
                return false;
        }
        return true;
    }
    
    int bfs(){
        int temp;
        head = 0;
        tail = 1;
        q[head].step = 0;
        q[head].x = start;
        memset(flag, 0, sizeof(flag));
        while(head < tail){
            for(int i = 1; i <= 9; i += 2){
                temp = q[head].x / 10 * 10 + i;
                if(judge_prim(temp) && !flag[temp]){
                    if(temp == end)
                        return q[head].step + 1;
                    else{
                        q[tail].x = temp;
                        q[tail ++].step = q[head].step + 1;
                        flag[temp] = true;
                    }
                }
                else 
                   continue;
            }
    
            for(int i = 0; i <= 9; i ++){
                temp = q[head].x /100 * 100 + 10 * i + q[head].x % 10;
                if(judge_prim(temp) && !flag[temp]){
                    if(temp == end)
                        return q[head].step + 1;
                    else{
                        q[tail].x = temp;
                        q[tail ++].step = q[head].step + 1;
                        flag[temp] = true;
                    }
                }
                else 
                    continue;
            }
    
            for(int i = 0; i <= 9; i ++){
                temp  = q[head].x / 1000 * 1000 + q[head].x / 10 % 10 * 10 + q[head].x % 10 + i * 100;
                if(judge_prim(temp) && !flag[temp]){
                    if(temp == end)
                        return q[head].step + 1;
                    else{
                        q[tail].x = temp;
                        q[tail ++].step = q[head].step + 1;
                        flag[temp] = true;
                    }
                }
                else 
                    continue;
            }
    
            for(int i = 1; i <= 9; i ++){
                temp = q[head].x % 1000 + i * 1000;
                if(judge_prim(temp) && !flag[temp]){
                    if(temp == end)
                        return q[head].step + 1;
                      else{
                      q[tail].x = temp;
                      q[tail ++].step = q[head].step + 1;
                      flag[temp] = true;
                    }
                }
            else 
                continue;
            }
            head ++;
        }
        return -1;
    }
    
    int main(void){
        cin >> T;
        int steps;
        while(T --){
            cin >> start >> end;
            if(start == end){
                cout << "0" << endl;
                continue;  
             }
            steps = bfs();
            if(steps != -1){
                cout << steps << endl;
                continue;
            }
            else 
                cout << "IMPOSSIBLE" << endl;
        }
        return 0;
    }


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  • 原文地址:https://www.cnblogs.com/chilumanxi/p/5136099.html
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