[Wikioi] 2495 jingting dance steps (A * + iterative deepening search)

 

I still want to explain this question. (The search is too weak. I did not think of a * at all. I was wrong when I was reading questions ,., --)

Well, the valuation is still that simple. You can determine the number of different colors (the UNICOM block in the upper left corner is not included)

Then a * is the same.

Iterative enhancement is still the same taste ~

Here, we use C [I] [J] to represent the outer layer of the UNICOM block and the UNICOM block starting in the upper left corner (because we want to expand the color from the outer layer), respectively marked as 1 and 2.

When searching, we only need to dye the color type of C [I] [J] to 2 and update the UNICOM block (no picture is required here, because the layer-by-layer expansion will increase monotonically, so we cannot use the previous color.) We only need to maintain the previous C array before searching, and then restore it after updating.

#include <cstdio>#include <cstring>#include <cmath>#include <string>#include <iostream>#include <algorithm>using namespace std;#define rep(i, n) for(int i=0; i<(n); ++i)#define for1(i,a,n) for(int i=(a);i<=(n);++i)#define for2(i,a,n) for(int i=(a);i<(n);++i)#define for3(i,a,n) for(int i=(a);i>=(n);--i)#define for4(i,a,n) for(int i=(a);i>(n);--i)#define CC(i,a) memset(i,a,sizeof(i))#define max(a,b) ((a)>(b)?(a):(b))#define min(a,b) ((a)<(b)?(a):(b))#define read(a) a=getnum()#define print(a) printf("%d", a)inline int getnum() { int ret=0; char c; int k=1; for(c=getchar(); c<‘0‘ || c>‘9‘; c=getchar()) if(c==‘-‘) k=-1; for(; c>=‘0‘ && c<=‘9‘; c=getchar()) ret=ret*10+c-‘0‘; return k*ret; }const int N=8;int m[N][N], c[N][N], n, ans;int fx[4]={-1, 1, 0, 0}, fy[4]={0, 0, -1, 1};bool used[6];inline int H() {int ret=0;CC(used, 0);rep(i, n) rep(j, n)if(!used[m[i][j]] && c[i][j]!=1)used[m[i][j]]=true, ++ret;return ret;}void change2(const int &x, const int &y, const int &color) {c[x][y]=1;int nx, ny;rep(i, 4) {nx=x+fx[i], ny=y+fy[i];if(nx<0 || nx>=n || ny<0 || ny>=n || c[nx][ny]==1) continue;c[nx][ny]=2;if(m[nx][ny]==color) change2(nx, ny, color);}}inline bool change(const int &color) {bool ret=false;rep(i, n) rep(j, n) if(c[i][j]==2 && m[i][j]==color) ret=true, change2(i, j, color);return ret;}bool dfs(const int &g) {int h=H();if(g+h>ans) return false;if(!h) return true;int t[N][N];memcpy(t, c, sizeof(c));for1(i, 0, 5) {if(change(i) && dfs(g+1)) return true;memcpy(c, t, sizeof(t));}return false;}int main() {for(read(n); n; read(n)) {CC(c, 0); CC(m, 0);rep(i, n) rep(j, n) read(m[i][j]);change2(0, 0, m[0][0]);for(ans=0; ; ++ans)if(dfs(0)) break;printf("%d\n", ans);}return 0;}