T1 rotator segment
This question is. .
My head is filled with a little bit of nature did not see it.
The most basic nature of the program is to reach every point as a fixed point. What is the same program? It is only by operating a reach, it actually refers to a fixed center of rotation. I thought of it, got the 60 points.
Consider optimization, this approach where the problem will happen? That is, every time you want to enumerate enumeration rotation point range statistics answer, but in the process of enumeration range and there will be a lot of redundant computations, so we can look to record a point for the next axis, extended to both sides, only ans statistical change the answer, and maintain two prefixes, a rotational number of the front fixing point statistics, and the other represents rotated.
Brother Fucai.
1 #include<cstdio> 2 #include<algorithm> 3 #include<vector> 4 #define HZOI std 5 using namespace HZOI; 6 const int N=1e6+5; 7 int n,res,ans; 8 int a[N],b[N],sm[N<<1]; 9 vector<int> vec[N<<1]; 10 inline void read(int &); 11 inline int max(int a,int b) { return a>b?a:b; } 12 int main() 13 { 14 int flag=0; 15 read(n); 16 for (int i=1; i<=n; ++i) 17 { 18 read(a[i]); 19 if (a[i]==i) b[2*i]=1,++res; 20 b[i*2]+=b[i*2-2]; 21 ++sm[a[i]+i]; 22 vec[a[i]+i].push_back(abs(a[i]*2-(a[i]+i))); 23 } 24 for (int i=1; i<=n*2; ++i) 25 { 26 sort(vec[i].begin(),vec[i].end(),greater<int>()); 27 for (int j=0; j<vec[i].size(); ++j) 28 { 29 int r=vec[i][j]; 30 ans=max(ans,b[n<<1]-b[i+r]+b[i-r-2]+sm[i]); 31 --sm[i]; 32 } 33 } 34 printf("%d",ans); 35 } 36 inline void read(int &a) 37 { 38 char ch=getchar(); 39 for (a=0; ch<'0' or ch>'9';ch=getchar()); 40 for (;ch>='0' and ch<='9'; a=(a<<3)+(a<<1)+(ch^48),ch=getchar()); 41 }
T2 go plaid
Large analog, but not purely analog.
The idea is very good, very strange.
For each point, it may lead to the up, down, left, and right grid (if the white grid), if you want to transfer, and if you must come to a wall, re-transmission. The question is required in the shortest distance, rather than that position is reached, which is different from simulation, we think the most short-circuit.
Along the shortest think the idea down, how to build side? I think of a point, even with up to eight point edge (up, down, left, right, on the wall, the wall, the left wall and right wall), then how to build map. For a point i want to transfer to the far point, it must first go at a wall, then we had a problem: the nearest wall location is not good demand. We thought that since we use the shortest path, as long as each is optimal from the point on the line (meaning that regardless of contribution went to other points, so the weight will be counted, and not ask), we consider the most topical optimal solution, and that is a point to go in a straight line, it went to the nearest wall, and then transferred to another three wall by this wall.
However, this can only be regarded Ideally yy out, you want to deal with that wall is a problem, the specific operation, because there is Dij, we do not consider that wall and then spread to come in if it takes itself (although it is legal , but it can effectively reduce the amount of code), that is to say, that will come closest point and other points equally, both from +1, which makes the answer will not make another mistake, after all, is the most short-circuit evaluation, and I just each point in itself guarantee optimal on it.
n ^ 2 treatment nearest building and construction side and then run Dij, resolved.
Brother Fucai.
1 #include<cstdio> 2 #include<cstring> 3 #include<vector> 4 #include<queue> 5 #include<algorithm> 6 #define HZOI std 7 using namespace HZOI; 8 const int N=555; 9 inline void read(int &); 10 struct DIJ{ 11 int num,dis; 12 friend bool operator < (DIJ a,DIJ b) 13 { 14 return a.dis>b.dis; 15 } 16 }poi[N*N<<1]; 17 int n,m; 18 int mp[N][N],sx,sy,tx,ty; 19 int ed[4][N*N],len[4][N*N];//左、上、右、下 20 int tt,first[N*N],vv[N*N<<5],nx[N*N<<5],ww[N*N<<5]; 21 int v[N*N]; 22 int stpx[4]={-1,1,0,0},stpy[4]={0,0,-1,1}; 23 void Dij(int ); 24 inline void Add(int ,int ,int ); 25 int main() 26 { 27 read(n),read(m); 28 for (int i=1; i<=n; ++i) 29 for (int j=1; j<=m; ++j) 30 { 31 char ch=getchar(); 32 while (ch^'.' and ch^'#' and ch^'C' and ch^'F') ch=getchar(); 33 if (ch=='C') sx=i,sy=j; 34 else if (ch=='F') tx=i,ty=j; 35 else if (ch=='#') mp[i][j]=1; 36 } 37 for (int i=1,d; i<=n; ++i) 38 for (int j=1; j<=m; ++j) 39 { 40 int id=(i-1)*m+j; 41 if (mp[i][j]) continue; 42 if (mp[i][j-1]) ed[0][id]=id,len[0][id]=0; 43 else ed[0][id]=ed[0][id-1],len[0][id]=len[0][id-1]+1; 44 if (mp[i-1][j]) ed[1][id]=id,len[1][id]=0; 45 else ed[1][id]=ed[1][(i-2)*m+j],len[1][id]=len[1][(i-2)*m+j]+1; 46 } 47 for (int i=n,d; i; --i) 48 for (int j=m; j; --j) 49 { 50 int id=(i-1)*m+j; 51 if (mp[i][j]) continue; 52 if (mp[i][j+1]) ed[2][id]=id,len[2][id]=0; 53 else ed[2][id]=ed[2][id+1],len[2][id]=len[2][id+1]+1; 54 if (mp[i+1][j]) ed[3][id]=id,len[3][id]=0; 55 else ed[3][id]=ed[3][i*m+j],len[3][id]=len[3][i*m+j]+1; 56 } 57 for (int i=1; i<=n; ++i) 58 for (int j=1; j<=m; ++j) 59 { 60 if (mp[i][j]) continue; 61 int id=(i-1)*m+j,minn=0x3f3f3f3f ; 62 for ( int k = 0 ; k < 4 ; ++ k) 63 { 64 if (! Mp [i + stpx [k]] [j + stpy [k]]) 65 Add (id (i + stpx [k ] - 1 ) * m + j + stpy [k], 1 ); 66 minn = min (len [k] [id], memory); 67 } 68 for ( int k = 0 ; k < 4 ; ++ k) 69 if (minn == len [k] [id] and len [k] [id] ^ 0 ) 70 Add(id,ed[k][id],minn); 71 else if (minn^len[k][id]) 72 Add(id,ed[k][id],minn+1); 73 } 74 memset(v,0,sizeof(v)); 75 Dij((sx-1)*m+sy); 76 if (poi[(tx-1)*m+ty].dis==0x3f3f3f3f) puts("no"); 77 else printf("%d\n",poi[(tx-1)*m+ty].dis); 78 return 0; 79 } 80 void Dij(int k) 81 { 82 priority_queue<DIJ> q; 83 for (int i=1; i<=n; ++i) 84 for (int j=1; j<=m; ++j) 85 poi[(i-1)*m+j].dis=0x3f3f3f3f, 86 poi[(i-1)*m+j].num=(i-1)*m+j; 87 poi[k].dis=0; 88 poi[k].num=k; 89 q.push(poi[k]); 90 while (!q.empty()) 91 { 92 DIJ x=q.top(); 93 q.pop(); 94 int pnt=x.num,dtc=x.dis; 95 v[pnt]=1; 96 for (int i=first[pnt]; i; i=nx[i]) 97 { 98 int ver=vv[i]; 99 if (v[ver]) continue; 100 if (poi[ver].dis>dtc+ww[i]) 101 { 102 poi[ver].dis=dtc+ww[i]; 103 q.push(poi[ver]); 104 } 105 } 106 } 107 } 108 inline void Add(int u,int v,int w) 109 { 110 vv[++tt]=v,ww[tt]=w,nx[tt]=first[u],first[u]=tt; 111 } 112 inline void read(int &a) 113 { 114 char ch=getchar(); 115 for (a=0; ch<'0' or ch>'9';ch=getchar()); 116 for (;ch>='0' and ch<='9'; a=(a<<3)+(a<<1)+(ch^48),ch=getchar()); 117 }
T3 histogram
Immortal question, magical number like array.
Will a little bit, but still hung Titans blog it.
Also experience a metaphysics algorithms, simulated annealing .
1 #include<cstdio> 2 #include<cstdlib> 3 #include<cmath> 4 #include<ctime> 5 #include<algorithm> 6 #define int long long 7 #define HZOI std 8 using namespace HZOI; 9 const int N=1e5+5,INF=1<<30; 10 double T=0.001,delta=0.97,eps=1e-10; 11 int n,ans,a[N],b[N],nowpos,nowans,newpos,newans; 12 inline int Ask(int ); 13 signed main() 14 { 15 srand(time(0)); 16 ans=1e16; 17 scanf("%lld",&n); 18 for (int i=1; i<=n; ++i) 19 scanf("%lld",&a[i]); 20 nowpos=(n+1)>>1; 21 nowans=Ask(nowpos); 22 while (T>eps) 23 { 24 // printf("T=%.12lf\n",T); 25 newpos=max(1ll,newpos); 26 newpos=min(n,newpos); 27 newans=Ask(newpos); 28 if( newans<nowans or exp((newans-nowans)/T)*RAND_MAX<rand() ) 29 nowpos=newpos,nowans=newans; 30 if (newans<ans) ans=newans; 31 T*=delta; 32 } 33 printf("%lld",ans); 34 } 35 inline int Ask(int k) 36 { 37 int t=1,sm=0,mid=(n+1)>>1,maxx; 38 b[k]=a[k]; 39 for (int i=k-1; i; --i) 40 b[i]=a[i]+t,++t; 41 t=1; 42 for (int i=k+1; i<=n; ++i) 43 b[i]=a[i]+t,++t; 44 nth_element(b+1,b+mid,b+n+1); 45 maxx=max(b[mid],max(k,n-k+1)); 46 for (int i=1; i<=n; ++i) sm+=abs(maxx-b[i]); 47 return sm; 48 }