USACO 5.1 Starry Night

Starry Night
IOI 98

High up in the night sky, the shining stars appear in clusters of various shapes. A cluster is a non-empty group of neighbouring stars, adjacent in horizontal, vertical or diagonal direction. A cluster cannot be a part of a larger cluster.

Clusters may be similar. Two clusters are similar if they have the same shape and number of stars, irrespective of their orientation. In general, the number of possible orientations for a cluster is eight, as Figure 1 exemplifies.

Figure 1. Eight similar clusters

The night sky is represented by a sky map, which is a two-dimensional matrix of 0's and 1's. A cell contains the digit 1 if it has a star, and the digit 0 otherwise.

Given a sky map, mark all the clusters with lower case letters. Similar clusters must be marked with the same letter; non-similar clusters must be marked with different letters.

You mark a cluster with a lower case letter by replacing every 1 in the cluster by that lower case letter.

PROGRAM NAME: starry

INPUT FORMAT

The first two lines contain, respectively, the width W and the height H of a sky map. The sky map is given in the following H lines, of W characters each.

SAMPLE INPUT (file starry.in)

23 15 10001000000000010000000 01111100011111000101101 01000000010001000111111 00000000010101000101111 00000111010001000000000 00001001011111000000000 10000001000000000000000 00101000000111110010000 00001000000100010011111 00000001110101010100010 00000100110100010000000 00010001110111110000000 00100001110000000100000 00001000100001000100101 00000001110001000111000In this case, the sky map has width 23 and height 15. Just to make it clearer, notice that this input file corresponds to the following picture of the sky.

Figure 2. Picture of the sky

OUTPUT FORMAT

The output file contains the same map as the input file, except that the clusters are marked as described in Task.

There will generally be more than one way to label the clusters with letters. Your program should choose the labeling such that if the entire output file is read as a string, this string will be minimal in the lexicographical ordering.

SAMPLE OUTPUT (file starry.out)

a000a0000000000b0000000 0aaaaa000ccccc000d0dd0d 0a0000000c000c000dddddd 000000000c0b0c000d0dddd 00000eee0c000c000000000 0000e00e0ccccc000000000 b000000e000000000000000 00b0f000000ccccc00a0000 0000f000000c000c00aaaaa 0000000ddd0c0b0c0a000a0 00000b00dd0c000c0000000 000g000ddd0ccccc0000000 00g0000ddd0000000e00000 0000b000d0000f000e00e0b 0000000ddd000f000eee000This is one possible result for the sample input above. Notice that this output file corresponds to the following picture.

Figure 3. Picture with the clusters marked

Constraints

0 <= W (width of the sky map) <= 100
0 <= H (height of the sky map) <= 100
0 <= Number of clusters <= 500
0 <= Number of non-similar clusters <= 26 (a..z)
1 <= Number of stars per cluster <= 160

————————————————————————————题解

这道题就是个暴力模拟题

除了恶心人没有别的作用

对称轴我们可以选成50

旋转就是把行的序号=列的序号，列的序号=100-行的序号+1

相似维护就是简单的并查集

真的好恶心啊qwq但是过了还是有一点成就感的，拿最后的数据输出了一下给自己看了看图像的样子是什么

 /*

 ID: ivorysi

 LANG: C++

 TASK: starry

 */

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 #include <algorithm>

 #include <queue>

 #include <set>

 #include <vector>

 #include <string.h>

 #define siji(i,x,y) for(int i=(x);i<=(y);++i)

 #define gongzi(j,x,y) for(int j=(x);j>=(y);--j)

 #define xiaosiji(i,x,y) for(int i=(x);i<(y);++i)

 #define sigongzi(j,x,y) for(int j=(x);j>(y);--j)

 #define inf 0x7fffffff

 #define ivorysi

 #define mo 97797977

 #define hash 974711

 #define base 47

 #define pss pair<string,string>

 #define MAXN 30005

 #define fi first

 #define se second

 #define pii pair<int,int>

 using namespace std;

 char m[][];

 int maze[][],id[][];

 int w,h,cnt,fa[],color[],tot,used[];

 int dix[]={,-,,,,-,,-};

 int diy[]={,,,-,,,-,-};

 vector<pii > v[];

 pii size[],poi[];

 int getfa(int x) {

     return fa[x]==x ? x : fa[x]=getfa(fa[x]);

 }

 int to[],bo[],lf[],ri[];

 void dfs(int x,int y) {

     id[x][y]=cnt;

     if(y<lf[cnt]) lf[cnt]=y;

     if(y>ri[cnt]) ri[cnt]=y;

     if(x<to[cnt]) to[cnt]=x;

     if(x>bo[cnt]) bo[cnt]=x;

      v[cnt].push_back(make_pair(x,y));

     xiaosiji(i,,) {

         if(maze[x+dix[i]][y+diy[i]] && id[x+dix[i]][y+diy[i]]==) {

             dfs(x+dix[i],y+diy[i]);

         }

     }

 }

 bool circ(int ori,int ch) {

     pii t=make_pair(,);

     xiaosiji(i,,v[ori].size()) {

         int temp=v[ori][i].fi;

         v[ori][i].fi=v[ori][i].se;

         v[ori][i].se=-temp+;

         if(v[ori][i]>t) t=v[ori][i];

     }

     int inx=poi[ch].fi-t.fi,iny=poi[ch].se-t.se;

     xiaosiji(i,,v[ori].size()) {

         v[ori][i].fi+=inx;

         v[ori][i].se+=iny;

     }

     sort(v[ori].begin(),v[ori].end());

     xiaosiji(i,,v[ori].size()) {

         if(v[ori][i]!=v[ch][i]) {

             return false;

         }

     }

     return true;

 }

 void sym(int ori) {

     xiaosiji(i,,v[ori].size()) {

         v[ori][i].se=-v[ori][i].se;

     }

 }

 void solve() {

     scanf("%d%d",&w,&h);

     siji(i,,h) {

         scanf("%s",m[i]+);

     }

     siji(i,,h) {

         siji(j,,w) {

             maze[i][j]=m[i][j]-'';

         }

     }

     memset(to,,sizeof(to));

     memset(lf,,sizeof(lf));

     siji(i,,h){

         siji(j,,w) {

             if(maze[i][j]) {

                 if(id[i][j]==) {

                     v[++cnt].clear();

                     dfs(i,j);

                     size[cnt]=make_pair(bo[cnt]-to[cnt],ri[cnt]-lf[cnt]);

                     if(size[cnt].fi > size[cnt].se)

                         swap(size[cnt].fi,size[cnt].se);

                 }

             }

         }

     }

     siji(i,,cnt) fa[i]=i;

     siji(i,,cnt) {

         sort(v[i].begin(),v[i].end());

         pii t=make_pair(,);

         xiaosiji(j,,v[i].size()) {

             if(v[i][j]>t) t=v[i][j];

         }

         poi[i]=t;

         siji(j,i+,cnt) {

             if(getfa(i)!=getfa(j)) {

                 if(v[j].size() != v[i].size()) {continue;}

                 if(size[i] != size[j]) {continue;}

                 siji(k,,){

                     if(circ(j,i)) {fa[getfa(j)]=getfa(i);break;}

                 }

                 sym(j);

                 siji(k,,){

                     if(circ(j,i)) {fa[getfa(j)]=getfa(i);break;}

                 }

             }

         }

     }

     siji(i,,cnt) {

         if(!used[getfa(i)]) {

             used[getfa(i)]=;

             color[getfa(i)]=++tot;

         }

         else {

             color[i]=color[getfa(i)];

         }

     }

     siji(i,,h) {

         siji(j,,w) {

             if(id[i][j]!=) {

                 printf("%c",color[id[i][j]]+'a'-);

             }

             else {printf("");}

         }

         puts("");

     }

 }

 int main(int argc, char const *argv[])

 {

 #ifdef ivorysi

     freopen("starry.in","r",stdin);

     freopen("starry.out","w",stdout);

 #else

     freopen("f1.in","r",stdin);

 #endif

     solve();

     return ;

 }