标签:clu 现在 time ini git soft nec ios inf
[POJ]P3126
Prime Path
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 35230 | Accepted: 18966 |
Description
1033The 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.
1733
3733
3739
3779
8779
8179
Input
Output
Sample Input
3 1033 8179 1373 8017 1033 1033
Sample Output
6 7 0
Source
#include<iostream> #include<cstdio> #include<algorithm> #include<cstring> using namespace std; const int Maxn = 10010 ; const int inf = 100000000 ; namespace iNx{ struct qq{ int num; int c; }; qq q[Maxn]; int primer[Maxn],pos[Maxn]; bool check[Maxn],exist[Maxn]; int cnt,best; void Euler(){ memset(check,1,sizeof check); int i,j; check[1]=false ; for(i=2;i<10000;i++){ if(check[i]) primer[++cnt]=i; for(j=1;j<=cnt&&(i*primer[j]<10000);j++){ check[i*primer[j]]=false ; if(i%primer[j]==0) break ; } } } int getn(int a,int i){ if(i==1) return a%10; if(i==2) return (a/10)%10; if(i==3) return (a/100)%10; return a/1000; } void bfs(int a,int b){ int head=1,tail=1; q[1].num=a; q[1].c=0; int i,j,h,k,m,t; while(tail>=head){ h=q[head].num; if(h==b){ printf("%d\n",q[head].c); break ; } exist[h]=true ; for(i=1;i<=4;i++){ k=getn(h,i); for(j=1,m=1;j<i;j++) m*=10; for(j=0;j<=9;j++){ if(i==4&&j==0) continue ; if(j==k) continue ; t=h+j*m-k*m; if(check[t]&&(!exist[t])){ q[++tail].num=t; q[tail].c=q[head].c+1; } } } head++; } } int main(){ Euler(); int n,a,b,i; scanf("%d",&n); for(i=1;i<=n;i++){ scanf("%d%d",&a,&b); memset(exist,0,sizeof exist); bfs(a,b); } return 0; } } int main(){ iNx::main(); return 0; }
现在要开始天天练习了,蒟蒻我太难了。
标签:clu 现在 time ini git soft nec ios inf
原文地址:https://www.cnblogs.com/Liisa/p/11478805.html