POJ 3126 Prime Path
Portal:http://poj.org/problem?id=3126
NWERC2006的题
BFS水题 一道好题
POJ16ms过,HDU0ms过,开心
有很多个坑,写了vis以后 复杂度就降低到常数级了,操作次数大概在1e4*1e3,所以可以乱搞
1 #include<iostream> 2 #include<algorithm> 3 #include<set> 4 #include<cstdio> 5 #include<cstdlib> 6 #include<cmath> 7 using namespace std; 8 #define FOR(i,j,k) for(int i=j;i<=k;i++) 9 #define FORD(i,j,k) for(int i=j;i>=k;i--) 10 #define LL long long 11 #define SZ(x) x.size() 12 int a[10010]; 13 void is_prime(int x) 14 { 15 FOR(i,3,int(sqrt(x))) 16 { 17 if(!(x%i)) return; 18 } 19 a[x]=1; 20 return; 21 } 22 int main() 23 { 24 int sum=0; 25 freopen("1.in","r",stdin); 26 freopen("1.out","w",stdout); 27 for(int i=1001;i<=9999;i+=2) 28 is_prime(i); 29 FOR(i,1001,9999) 30 if(a[i]){cout<<i<<',';sum++;} 31 cout<<sum; 32 fclose(stdin); 33 fclose(stdout); 34 return 0; 35 }
1 #include<iostream> 2 #include<algorithm> 3 #include<set> 4 #include<cstdio> 5 #include<cstdlib> 6 #include<cmath> 7 #include<queue> 8 using namespace std; 9 #define FOR(i,j,k) for(int i=j;i<=k;i++) 10 #define FORD(i,j,k) for(int i=j;i>=k;i--) 11 #define LL long long 12 #define SZ(x) x.size() 13 int 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9433,9437,9439,9461,9463,9467,9473,9479,9491,9497,9511,9521,9533,9539,9547,9551,9587,9601,9613,9619,9623,9629,9631,9643,9649,9661,9677,9679,9689,9697,9719,9721,9733,9739,9743,9749,9767,9769,9781,9787,9791,9803,9811,9817,9829,9833,9839,9851,9857,9859,9871,9883,9887,9901,9907,9923,9929,9931,9941,9949,9967,9973}; 14 int p[10000],vis[10000]; 15 int k,n,m,s,ans; 16 int bfs() 17 { 18 queue<int> q1,q2; 19 q1.push(n); 20 q2.push(1); 21 while(q1.front()!=m) 22 { 23 int fr=q1.front(); 24 int fv=q2.front(); 25 int shi=1; 26 vis[fr]=1; 27 FOR(i,1,4) 28 { 29 ans=fr/(shi*10)*(shi*10)+fr%shi; 30 if(i!=4) s=0; else s=1; 31 FOR(j,s,9) if(p[ans+shi*j]) if(!vis[ans+shi*j]) {q1.push(ans+shi*j); q2.push(fv+1);vis[ans+shi*j]=1;} 32 shi*=10; 33 } 34 q1.pop(); 35 q2.pop(); 36 } 37 return q2.front()-1; 38 } 39 int main() 40 { 41 FOR(i,0,1060) 42 p[a[i]]=1; 43 cin>>k; 44 FOR(i,1,k) 45 { 46 cin>>n>>m; 47 FOR(i,1000,9999) 48 vis[i]=0; 49 cout<<bfs()<<endl; 50 } 51 return 0; 52 }
枚举位
fr/(shi*10)*(shi*10)+shi*j+fr%shi;
int shi=1; FOR(i,1,4) { ans=fr/(shi*10)*(shi*10)+fr%shi; if(i!=4) s=0; else s=1; FOR(j,s,9) if(p[ans+shi*j]) if(!vis[ans+shi*j]) {q1.push(ans+shi*j); q2.push(fv+1);vis[ans+shi*j]=1;} shi*=10; }
好孩子不要学我,应该在while处判断q.empty() 否则自测时可能死循环
其他版本:
HDU | 1973 | |||
POJ | 3126 | |||
SPOJ | PPATH | |||
UVA | 12101 | |||
UVALive | 3639 |