题目(https://vjudge.net/problem/CodeForces-520B)OC
大致意思 :有两个数m,n,执行乘2或者减1的操作(m不能小于0),问要得到n所要执行的最少操作次数
//bfs的话注意边界的now.num<n*2,别的就是模板了
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<string.h>
#include<cstdlib>
#include<cstring>
#include<math.h>
#include<cmath>
#include<vector>
#include<map>
#include<queue>
using namespace std;
struct node
{
int num;
int step;
};
queue<node>q;
int m,n;
int book[100001];
int bfs()
{
node f;
f.num=m;
book[m]=1;
f.step=0;
q.push(f);
while(!q.empty())
{
node top=q.front();
node now;
q.pop();
if(top.num==n)
return top.step;
for(int i=0;i<2;i++)
{
if(i==0)
{
now.num=top.num-1;
if(now.num>0&&now.num<n*2&&book[now.num]==0)
{
now.step=top.step+1;
book[now.num]=1;
q.push(now);
}
}
if(i==1)
{
now.num=top.num*2;
if(now.num>0&&now.num<n*2&&book[now.num]==0)
{
now.step=top.step+1;
book[now.num]=1;
q.push(now);
}
}
}
}
}
int main()
{
scanf("%d%d",&m,&n);
memset(book,0,sizeof(book));
if(m>=n)
{
printf("%d\n",m-n);
}
else
printf("%d\n",bfs());
return 0;
}
- dfs的话刚开始写一直WA,后来一直卡在test9,11,应该是数字大了10000,数组没开大,溢出了,**然后是memset给数组a赋初值的时候,不能赋值为无穷大,改成fill函数(说实话,fill函数也蛮好用的QAQ)
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<string.h>
#include**<cstdlib>
#include<cstring>
#include<math.h>
#include<cmath>
#include<vector>
#include<map>
#include<queue>
using namespace std;
int m,n;
int a[30001];
void dfs(int x,int step)
{
if(x<=0)
return;
if(a[x]<step)//剪枝,这次的step与上条到达该出的路径长度比较
return;
a[x]=step;
if(x>n)
{
dfs(n,step+x-n);
return;
}
if(x==n)
{
a[x]=min(step,a[x]);//这次的step与上一次标记a[x]比较取小的
return ;
}
dfs(2*x,step+1);
dfs(x-1,step+1);
}
int main()
{
scanf("%d%d",&m,&n);
fill(a,a+30001,999999);
a[m]=0;
dfs(m,0);
cout<<a[n]<<endl;
return 0;
}
- 然后是逆推了,这段代码比dfs还简单,如果结果是偶数,那一定是偶数/2变过来的,反过来*2,如果是奇数,就是-1变过来的,反过来加上1,当操作过程遇到m>n时,一步一步减1
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<string.h>
#include<cstdlib>
#include<cstring>
#include<math.h>
#include<cmath>
#include<vector>
#include<map>
#include<queue>
using namespace std;
int main()
{
int m,n;
scanf("%d%d",&m,&n);
int ans=0;
while(m<n)
{
if(n%2==0)
{
n/=2;
}
else if(n%2==1)
{
n++;
}
ans++;
}
printf("%d\n",ans+m-n);
return 0;
}