poj 3278

来源：互联网发布：php pack n 编辑：程序博客网时间：2024/05/14 10:44

这是一道很水的bfs 题目。点击打开链接

但是要注意的是数组要开大点。因为可能会访问到200000.

注意剪枝。否则会超时的。

219ms

#include <iostream>#include <memory.h>#include <queue>using namespace std;const int N=200010;int n,k,ans;bool vis[N];int cnt[N];void bfs(int n,int k){    memset(vis,false,sizeof(vis));    memset(cnt,0,sizeof(cnt));    queue<int > q;    int cur,tmp;    vis[n]=true;    cnt[n]=0;    q.push(n);    while(!q.empty())    {        cur=q.front();        q.pop();        if(cur==k)        {            ans=cnt[cur];            return ;        }        for(int i=1;i<=3;i++)        {            if(i==1)            {                if(cur>100000)                    continue;                tmp=cur+1;            }            else if(i==2)            {                if(cur==0)                    continue;                tmp=cur-1;            }            else            {                if(cur>66667)                    continue;                tmp=2*cur;            }            if(!vis[tmp])            {                cnt[tmp]=cnt[cur]+1;                vis[tmp]=true;                q.push(tmp);            }        }    }}int main(){    while(cin>>n>>k)    {        bfs(n,k);        cout<<ans<<endl;    }}

//通过修改主函数里面

/*if(abs(n-k) <=1)   {  cout<<abs(n-k)<<endl;  continue;

       if((n << 1) == k){cout<<1<<endl;continue;}        if(k < n){cout<<n-k<<endl;continue;}        bfs(n,k);        cout<<ans<<endl;

*/

时间就优化到了157ms

引用 poj上

Alex_Shepherd 的代码。双向bfs 太牛逼了。留着学习

32ms

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#include <string>#include <vector>#include <queue>#include <stack>#include <set>#include <map>#include <limits.h>#include <bitset>#define F "%I64d"using namespace std;typedef long long ll;int dp[2][200000],N,K;queue<int> QN,QK;int bfs(){while(!QN.empty())QN.pop();while(!QK.empty())QK.pop();memset(dp,-1,sizeof dp);QN.push(N);QK.push(K);dp[0][N] = dp[1][K] = 0;while(1){if(QN.front() > 1 && dp[0][QN.front() - 1] == -1){dp[0][QN.front() - 1] = dp[0][QN.front()] + 1;if(dp[1][QN.front() - 1] != -1)return dp[0][QN.front() - 1] + dp[1][QN.front() - 1];QN.push(QN.front() - 1);}if(QN.front() < 100000 && dp[0][QN.front() + 1] == -1){dp[0][QN.front() + 1] = dp[0][QN.front()] + 1;if(dp[1][QN.front() + 1] != -1)return dp[0][QN.front() + 1] + dp[1][QN.front() + 1];QN.push(QN.front() + 1);}if(QN.front() < 100000 && dp[0][QN.front() << 1] == -1){dp[0][QN.front() << 1] = dp[0][QN.front()] + 1;if(dp[1][QN.front() << 1] != -1)return dp[0][QN.front() << 1] + dp[1][QN.front() << 1];QN.push(QN.front() << 1);}if(QK.front() > 1 && dp[1][QK.front() - 1] == -1){dp[1][QK.front() - 1] = dp[1][QK.front()] + 1;if(dp[0][QK.front() - 1] != -1)return dp[1][QK.front() - 1] + dp[0][QK.front() - 1];QK.push(QK.front() - 1);}if(QK.front() < 100000 && dp[1][QK.front() + 1] == -1){dp[1][QK.front() + 1] = dp[1][QK.front()] + 1;if(dp[0][QK.front() + 1] != -1)return dp[1][QK.front() + 1] + dp[0][QK.front() + 1];QK.push(QK.front() + 1);}if((QK.front() & 1) == 0 && dp[1][QK.front() >> 1] == -1){dp[1][QK.front() >> 1] = dp[1][QK.front()] + 1;if(dp[0][QK.front() >> 1] != -1)return dp[1][QK.front() >> 1] + dp[0][QK.front() >> 1];QK.push(QK.front() >> 1);}QN.pop();QK.pop();}}int main(){N:while(~scanf("%d%d",&N,&K)){if(abs(N - K) <= 1){printf("%d\n",abs(N - K));continue;}if((N << 1) == K){puts("1");continue;}if(K < N){printf("%d\n",N - K);continue;}printf("%d\n",bfs());}return 0;}

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