题目链接
IDA*,估价函数为当前除了左上角的连通块以外颜色的种类数,因为每次最多消去一个颜色。
维护位于当前连通块的边缘但颜色不同的点,每次从这些点拓展就行。
#include <cstdio>
#include <cstring>
int a[10][10], n, l[] = {-1, 1, 0, 0}, r[] = {0, 0, -1, 1}, v[10], vis[10][10], limit;
void dfs(int now, int x, int y){
//printf("%d %d %d
", now, x, y);
vis[x][y] = 1;
for(int i = 0; i < 4; ++i){
int X = x + l[i], Y = y + r[i];
if(X < 0 || Y < 0 || X >= n || Y >= n || vis[X][Y] == 1) continue;
vis[X][Y] = 2;
if(a[X][Y] == now) dfs(now, X, Y);
}
}
int fill(int now){
int p = 0;
for(int i = 0; i < n; ++i)
for(int j = 0; j < n; ++j)
if(a[i][j] == now && vis[i][j] == 2)
dfs(now, i, j), ++p;
return p;
}
int IDA(int now, int dep){
v[0] = v[1] = v[2] = v[3] = v[4] = v[5] = 0;
for(int i = 0; i < n; ++i)
for(int j = 0; j < n; ++j)
if(vis[i][j] != 1)
v[a[i][j]] = 1;
int f = v[0] + v[1] + v[2] + v[3] + v[4] + v[5];
if(!f) return 1;
if(dep + f > limit) return 0;
int tmp[10][10];
memcpy(tmp, vis, sizeof vis);
for(int k = 0; k < 6; ++k){
if(k == now) continue;
if(fill(k) && IDA(k, dep + 1)) return 1;
memcpy(vis, tmp, sizeof vis);
}
return 0;
}
int main(){
while(scanf("%d", &n) && n){
for(int i = 0; i < n; ++i)
for(int j = 0; j < n; ++j)
scanf("%d", &a[i][j]);
memset(vis, 0, sizeof vis);
dfs(a[0][0], 0, 0);
for(limit = 0; ; ++limit){
if(IDA(a[0][0], 0)){
printf("%d
", limit);
break;
}
}
}
}