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| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 13813 | Accepted: 7796 |
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. 1033The 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.
1733
3733
3739
3779
8779
8179
Input
Output
Sample Input
3 1033 8179 1373 8017 1033 1033
Sample Output
6 7 0
题意是说,给定两个四位素数a b 问从a变换到b,最少需要变换几次.
变换的要求是,每次只能改变一个数字,而且中间过程得到的四位数也必须为素数.
因为提到最少变换几次,容易想到bfs,bfs第一次搜到的一定是最短步数.
先打个素数表
然后写个函数判断两个四位数有几位数字不同,如果只有一位,返回true,否则返回false
然后竟然wa了两次!
下表写错!
pri[k++]=i;是先给pri[k]赋值,再k++;
pri[++k]=i;才是先增加,再赋值.这个搞错了.所以wa了....sad
/************************************************************************* > File Name: code/2015summer/searching/F.cpp > Author: 111qqz > Email: rkz2013@126.com > Created Time: Fri 24 Jul 2015 01:16:23 AM CST ************************************************************************/ #include<iostream> #include<iomanip> #include<cstdio> #include<algorithm> #include<cmath> #include<cstring> #include<string> #include<map> #include<set> #include<queue> #include<vector> #include<stack> #define y0 abc111qqz #define y1 hust111qqz #define yn hez111qqz #define j1 cute111qqz #define tm crazy111qqz #define lr dying111qqz using namespace std; #define REP(i, n) for (int i=0;i<int(n);++i) typedef long long LL; typedef unsigned long long ULL; const int N =1E4+5; int pri[N],which[N]; int a,b,k; bool flag; int d[N]; bool prime(int x) { for ( int i = 2 ; i*i<=x ;i++ ) { if (x %i==0) return false; } return true; } bool ok (int x,int y) { if (d[y]!=-1) return false; int res = 0; //记录两个数不对应不相等的数字的个数 int xx=x,yy=y; while (x&&y) { if (x%10!=y%10) res++; x = x/10; y = y/10; } // if (res==1) cout<<"x:"<<xx<<" y:"<<yy<<endl; if (res==1) return true; return false; } void bfs() { queue<int>x; memset(d,-1,sizeof(d)); x.push(a); d[a]=0; while (!x.empty()) { int px = x.front(); // cout<<"px:"<<px<<endl; x.pop(); if (px==b) return; for ( int i = 0 ; i< k ; i++ ) { if (ok(px,pri[i])) { d[pri[i]]=d[px]+1; x.push(pri[i]); } } } } int main() { k =0; for ( int i = 1000; i <=9999; i++ ) { if (prime(i)) { pri[k++]=i; } } int T; // cout<<pri[0]<<endl; // cout<<pri[1]<<endl; cin>>T; while (T--) { cin>>a>>b; bfs(); if (d[b]==-1) { cout<<"Impossible"<<endl; } else { cout<<d[b]<<endl; } } return 0; }
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原文地址：http://www.cnblogs.com/111qqz/p/4674633.html
