poj 1084 Square Destroyer dlx解重复覆盖

来源：互联网发布：linux创建新用户和组编辑：程序博客网时间：2024/05/21 10:53

分析：

将问题转化为重复覆盖问题，DancingLink解决。

代码：

//poj 1084//sep9#include <iostream>using namespace std;const int maxN=10024;const int maxL=128;int L[maxN],R[maxN],U[maxN],D[maxN];int C[maxN],H[maxN];int S[maxN],A[maxN],X[maxN];bool makeup[maxL][maxL];int cnt,ans;void build(int r,int c){for(int i=0;i<=c;++i){U[i]=D[i]=i;L[i]=i==0?c:i-1;R[i]=(i+1)%(c+1);C[i]=i;//colum head's idS[i]=0;}for(int i=1;i<=r;++i)H[i]=-1;//row head's id;cnt=c;}void init(int n){memset(makeup,false,sizeof(makeup));int c=1;for(int len=1;len<=n;++len)for(int i=1;i+len<=n+1;++i)for(int j=1;j+len<=n+1;++j){for(int k=0;k<len;++k){makeup[(i-1)*(2*n+1)+j+k][c]=true;makeup[(i-1+len)*(2*n+1)+j+k][c]=true;makeup[i*n+(i-1)*(n+1)+j+k*(2*n+1)][c]=true;makeup[i*n+(i-1)*(n+1)+j+k*(2*n+1)+len][c]=true;}c++;}}void link(int r,int c){++cnt;C[cnt]=c;++S[c];X[cnt]=r;D[cnt]=D[c];U[D[c]]=cnt;D[c]=cnt;U[cnt]=c;if(H[r]<0)H[r]=L[cnt]=R[cnt]=cnt;else{R[cnt]=R[H[r]];L[cnt]=H[r];L[R[H[r]]]=cnt;R[H[r]]=cnt;}}void remove(int c){for(int i=D[c];i!=c;i=D[i]){L[R[i]]=L[i];R[L[i]]=R[i];}} void resume(int c){for(int i=D[c];i!=c;i=D[i]){L[R[i]]=i;R[L[i]]=i;}}int get_H(){bool has[maxL];memset(has,false,sizeof(has));int res=0;for(int i=R[0];i!=0;i=R[i])if(!has[i]){++res;for(int j=D[i];j!=i;j=D[j])for(int k=R[j];k!=j;k=R[k])has[C[k]]=true;}return res;}void dfs(int k){if(R[0]==0){ans=min(ans,k);return ;}else if(k+get_H()>=ans)return ;int c=R[0];for(int t=R[0],maxx=INT_MAX;t!=0;t=R[t])if(S[t]<maxx)maxx=S[t],c=t;for(int i=D[c];i!=c;i=D[i]){remove(i);for(int j=R[i];j!=i;j=R[j]){remove(j);--S[C[j]];}dfs(k+1);for(int j=L[i];j!=i;j=L[j]){resume(j);++S[C[j]];}resume(i);}}int main(){int cases,n,row,col;scanf("%d",&cases);while(cases--){scanf("%d",&n);col=0,row=2*n*(n+1);for(int i=1;i<=n;++i)col+=i*i;build(row,col);init(n);int num;scanf("%d",&num);bool vis[maxL];//mark square have been destoryedmemset(vis,false,sizeof(vis));while(num--){int r;scanf("%d",&r);for(int j=1;j<=col;++j)if(makeup[r][j]&&!vis[j]){vis[j]=true;R[L[j]]=R[j];L[R[j]]=L[j];}}for(int i=1;i<=row;++i)for(int j=1;j<=col;++j)if(makeup[i][j]&&!vis[j])link(i,j);ans=INT_MAX;dfs(0);printf("%d\n",ans);}return 0;}

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