广搜 poj3278 poj1426 poj3126

Catch That Cow

Time Limit: 2000 MS Memory Limit: 65536 KB

64-bit integer IO format: %I64d , %I64u Java class name: Main

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Description

Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.

* Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute
* Teleporting: FJ can move from any point X to the point 2 × X in a single minute.

If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?

Input

Line 1: Two space-separated integers: N and K

Output

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

Sample Input

5 17

Sample Output

4

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <queue>

using namespace std;
#define max 100002    ///看了别人的代码   说广搜的标记数组 会很大 如果开成局部的会PE
bool vis[max];
int deep[max];
int n,k;

int bfs()
{
    queue<int>q;
    q.push(n);
    memset(vis,false,sizeof(vis));  ///全没有遍历过
    vis[n]=true; ///起点遍历过了。。
    memset(deep,,sizeof(deep));  ///初始都定义为0  走了0步 17
    deep[n]=;

    while(!q.empty())
    {
        int tp=q.front();
        q.pop();
        vis[tp]=true;
        if(tp==k)
            break;
        else
        {
            int ans1=tp+;
            int ans2=tp-;
            int ans3=tp*;
            if(ans1<=&&!vis[ans1])
            {
                q.push(ans1);
                deep[ans1]=deep[tp]+;
                //cout<<deep[ans1]<<endl;
                vis[ans1]=true;
            }
            if(ans2>=&&!vis[ans2])   ///注意边界条件   刚刚>0  WA了
            {
                q.push(ans2);
                deep[ans2]=deep[tp]+;
                //cout<<deep[ans2]<<endl;
                vis[ans2]=true;

            }
            if(ans3<=&&!vis[ans3])
            {
                q.push(ans3);
                deep[ans3]=deep[tp]+;
               // cout<<deep[ans3]<<endl;
                vis[ans3]=true;

            }
        }
    }
    return deep[k];
}

int main()
{
   cin>>n>>k;
       printf("%d\n",bfs());
    return ;
}

第一道广搜题  参考别人的代码~~~~http://kymowind.blog.163.com/blog/static/1842222972011717112727688/

poj1426     http://blog.csdn.net/lyy289065406/article/details/6647917

Prime Path

Time Limit: 1000 MS Memory Limit: 65536 KB

64-bit integer IO format: %I64d , %I64u Java class name: Main

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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

3
1033 8179
1373 8017
1033 1033

Sample Output

6
7
0

题意：从第一个数编导第二个数最少需要多少步  要求：每次只能改变一个数且这个数与之前的数都不一样  同时这个数时素数 

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <queue>
#include <math.h>

using namespace std;
int prime[];

void inin()
{
    int i,j;
    for(i=;i<=;i++){
        for(j=;j<i;j++)
            if(i%j==){
                prime[i]=false;
                break;
            }
        if(j==i) prime[i]=true;
    }
}

int vis[]; ///是否遍历过
int t[]; ///存放各个位的数
int num[];  ///最后输出的值

int bfs(int n,int m)
{
    int temp;
    queue<int>q;
    q.push(n);
    memset(vis,,sizeof(vis));
    memset(num,,sizeof(num));
    vis[n]=;
    memset(t,,sizeof(t));

     while(!q.empty())
     {
         int tp=q.front();
         q.pop();
         if(tp==m)
         return num[tp];

        t[]=tp/;
        t[]=tp%/;
        t[]=tp%/;
        t[]=tp%;

         for(int i=;i<;i++)
         {
             temp=t[i];
             for(int j=;j<=;j++)
                 if(temp!=j)
                 {
                     t[i]=j;
                     int ans=t[]*+t[]*+t[]*+t[];

                     if(!vis[ans]&&prime[ans])
                     {
                         num[ans]=num[tp]+;
                         vis[ans]=;
                         q.push(ans);
                     }
                     if(ans==m)
                     return num[ans];
                 }
        t[i]=temp;
         }
     }
     return -;
}

int main()
{
    int n,m,tt;
    inin();
    cin>>tt;
    while(tt--)
    {
        cin>>n>>m;
        if(bfs(n,m)!=-)
        cout<<bfs(n,m)<<endl;
        else
        cout<<"Impossible"<<endl;
    }
    return ;
}

之前判断素数的函数不是这样写的  一直WA      可是我的错在哪里了呢  先贴上  以后看看

int prim(int n)
{
    for(int i=;i<sqrt(n);i++)
    {
        if(n%i==)
        {
            return false;
            break;
        }
    }
    return true;
}