标签：stat inpu ted i see ram com quit img pac
题目链接：
题目：
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 10737 | Accepted: 6110 |
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.
— 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.
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这道题目思想非常easy 。。可是实现起来比較复杂 。。
。
用bfs搜索。。
。
还有就是假设预处理打素数表的话会快许多。。
。
。可是我打的素数表很很的戳。。
。
。。可是还是0MS.。。
代码例如以下:
#include<cstdio>
#include<queue>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
const int maxn=9999+10;
int prime[maxn];
bool vis[maxn];
struct point
{
int x;
int time;
};
point start,end;
bool is_prime(int x)
{
for(int i=2;i<=sqrt(x*1.0);i++)
{
if(x%i==0)
return false;
}
return true;
}
void init()
{
for(int i=1000;i<=maxn;i++)
prime[i]=i;
for(int i=1000;i<=maxn;i++)
{
if(is_prime(prime[i]))
prime[i]=1;
else
prime[i]=0;
}
}
int bfs()
{
queue<point>q;
point old,current;
q.push(start);
while(!q.empty())
{
old=q.front();
q.pop();
if(old.x==end.x)
return old.time;
for(int i=1;i<=9;i=i+2)//个位
{
current.x=old.x/10*10+i;
if(current.x==old.x)
continue;
// printf("%d\n",current.time);
if(prime[current.x]&&!vis[current.x])
{
vis[current.x]=1;
current.time=old.time+1;
q.push(current);
}
}
for(int i=0;i<=9;i++)
{
current.x=old.x/100*100+i*10+old.x%10;
if(current.x==old.x)
continue;
// printf("%d\n",current.time);
if(prime[current.x]&&!vis[current.x])
{
vis[current.x]=1;
current.time=old.time+1;
q.push(current);
}
}
for(int i=0;i<=9;i++)
{
int tmp1=old.x%100;
current.x=(old.x/1000*10+i)*100+tmp1;
if(current.x==old.x)
continue;
if(prime[current.x]&&!vis[current.x])
{
vis[current.x]=1;
current.time=old.time+1;
q.push(current);
}
}
for(int i=1;i<=9;i++)
{
int tmp1=old.x%1000;
current.x=i*1000+tmp1;
if(current.x==old.x)
continue;
if(prime[current.x]&&!vis[current.x])
{
vis[current.x]=1;
current.time=old.time+1;//printf("%d\n",current.time);
q.push(current);
}
}
}
return -1;
}
int main()
{
memset(prime,0,sizeof(prime));
int T;
scanf("%d",&T);
init();
while(T--)
{
memset(vis,false,sizeof(vis));
scanf("%d %d",&start.x,&end.x);
start.time=0;
int ans=bfs();
//printf("nas:%d\n",ans);
if(ans!=-1)
printf("%d\n",ans);
else
printf("Impossible\n");
}
return 0;
}
标签：stat inpu ted i see ram com quit img pac
原文地址：http://www.cnblogs.com/wzjhoutai/p/7130349.html
