Catch That Cow 简单BFS
2012-09-26 22:29
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/*一个很基础的BFS。就是记录步数的时候,之前只用一个变量记录step是不行的。
#include <queue>
using namespace std;
const int maxn=200001;
int a[maxn];
bool vis[maxn];
int step[maxn];
int n,s,t;
void bfs(int s,int t)
{
queue <int> q;
q.push(s);
vis[s]=true;
while(!q.empty())
{
int u=q.front();
q.pop();
if(!vis[u+1]&&u+1>=0&&u+1<=maxn)
q.push(u+1),vis[u+1]=true,step[u+1]=step[u]+1;
if(u+1==t) return ;
if(!vis[u-1]&&u-1>=0&&u-1<=maxn)
q.push(u-1),vis[u-1]=true,step[u-1]=step[u]+1;
if(u-1==t) return ;
if(!vis[2*u]&&2*u>=0&&2*u<=maxn)
q.push(u*2),vis[2*u]=true,step[u*2]=step[u]+1;
if(u*2==t) return ;
}
}
int main()
{
scanf("%d%d",&s,&t);
memset(vis,false,sizeof(vis));
step[s]=0;
bfs(s,t);
printf("%d\n",step[t]);
return 0;
}
在每一部状态拓展的时候,要用一个数组来记录每一层转台拓展结束后的步数。 单用step记录的话会有很多重复的结果。*/ #include <stdio.h> #include <cstring> #include <queue> using namespace std; const int maxn=200001; int a[maxn]; bool vis[maxn]; int step[maxn]; int n,s,t; void bfs(int s,int t) { queue <int> q; q.push(s); vis[s]=true; while(!q.empty()) { int u=q.front(); q.pop(); if(!vis[u+1]&&u+1>=0&&u+1<=maxn) q.push(u+1),vis[u+1]=true,step[u+1]=step[u]+1; if(u+1==t) return ; if(!vis[u-1]&&u-1>=0&&u-1<=maxn) q.push(u-1),vis[u-1]=true,step[u-1]=step[u]+1; if(u-1==t) return ; if(!vis[2*u]&&2*u>=0&&2*u<=maxn) q.push(u*2),vis[2*u]=true,step[u*2]=step[u]+1; if(u*2==t) return ; } } int main() { scanf("%d%d",&s,&t); memset(vis,false,sizeof(vis)); step[s]=0; bfs(s,t); printf("%d\n",step[t]); return 0; }
#include <queue>
using namespace std;
const int maxn=200001;
int a[maxn];
bool vis[maxn];
int step[maxn];
int n,s,t;
void bfs(int s,int t)
{
queue <int> q;
q.push(s);
vis[s]=true;
while(!q.empty())
{
int u=q.front();
q.pop();
if(!vis[u+1]&&u+1>=0&&u+1<=maxn)
q.push(u+1),vis[u+1]=true,step[u+1]=step[u]+1;
if(u+1==t) return ;
if(!vis[u-1]&&u-1>=0&&u-1<=maxn)
q.push(u-1),vis[u-1]=true,step[u-1]=step[u]+1;
if(u-1==t) return ;
if(!vis[2*u]&&2*u>=0&&2*u<=maxn)
q.push(u*2),vis[2*u]=true,step[u*2]=step[u]+1;
if(u*2==t) return ;
}
}
int main()
{
scanf("%d%d",&s,&t);
memset(vis,false,sizeof(vis));
step[s]=0;
bfs(s,t);
printf("%d\n",step[t]);
return 0;
}
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