广搜 BFS POJ 3126 Prime Path
2016-07-28 17:30
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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
Sample Output
AC代码
#include<iostream>
#include<queue>
#include<cmath>
#include<cstring>
using namespace std;
queue<int> num;
int a,b,w[4],v[4],change[10010],vis[10010]; //w,v数组用于存改变后和改变前四位数的数字
bool isp[10010]; //生成素数表提高判断素数效率
bool ispreme(int q)
{
int m=sqrt(q);
for(int i=2;i<=m;i++)
if(q%i==0)
return false;
return true;
}
void bfs()
{
num.push(a);
vis[a]=1;
while(!num.empty()){
int u=num.front();
int temp=u;
for(int i=0;i<4;i++){ //更新改变后的四位数
w[i]=temp%10;
temp/=10;
}
num.pop();
int i,j;
for(i=0;i<4;i++){ //对四个位置分别改变一个数字
if(i!=3) j=0; //千位从1开始,其他从0开始替换
else j=1;
for(;j<10;j++){
int bk=u-w[i]*pow(10,i)+j*pow(10,i);
if(j!=w[i]&&isp[bk]&&!vis[bk]){
num.push(bk);
vis[bk]=1;
change[bk]=change[u]+1;
if(bk==b)
return;
}
}
}
}
}
int main()
{
int n;
for(int i=1000;i<10010;i++)
isp[i]=ispreme(i);
cin>>n;
while(n--){
cin>>a>>b;
if(a==b){
cout<<0<<endl;
continue;
}
while(!num.empty()) num.pop(); //初始化
memset(change,0,sizeof(change));
memset(vis,0,sizeof(vis));
int temp=b;
for(int i=0;i<4;i++){ //存储改变前的四位数字
v[i]=temp%10;
temp/=10;
}
bfs();
cout<<change[b]<<endl;
}
return 0;
}
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
AC代码
#include<iostream>
#include<queue>
#include<cmath>
#include<cstring>
using namespace std;
queue<int> num;
int a,b,w[4],v[4],change[10010],vis[10010]; //w,v数组用于存改变后和改变前四位数的数字
bool isp[10010]; //生成素数表提高判断素数效率
bool ispreme(int q)
{
int m=sqrt(q);
for(int i=2;i<=m;i++)
if(q%i==0)
return false;
return true;
}
void bfs()
{
num.push(a);
vis[a]=1;
while(!num.empty()){
int u=num.front();
int temp=u;
for(int i=0;i<4;i++){ //更新改变后的四位数
w[i]=temp%10;
temp/=10;
}
num.pop();
int i,j;
for(i=0;i<4;i++){ //对四个位置分别改变一个数字
if(i!=3) j=0; //千位从1开始,其他从0开始替换
else j=1;
for(;j<10;j++){
int bk=u-w[i]*pow(10,i)+j*pow(10,i);
if(j!=w[i]&&isp[bk]&&!vis[bk]){
num.push(bk);
vis[bk]=1;
change[bk]=change[u]+1;
if(bk==b)
return;
}
}
}
}
}
int main()
{
int n;
for(int i=1000;i<10010;i++)
isp[i]=ispreme(i);
cin>>n;
while(n--){
cin>>a>>b;
if(a==b){
cout<<0<<endl;
continue;
}
while(!num.empty()) num.pop(); //初始化
memset(change,0,sizeof(change));
memset(vis,0,sizeof(vis));
int temp=b;
for(int i=0;i<4;i++){ //存储改变前的四位数字
v[i]=temp%10;
temp/=10;
}
bfs();
cout<<change[b]<<endl;
}
return 0;
}
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