第五章 习题（二）

5-9 拉丁矩阵问题

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int NM=25;int x[NM][NM],n,m,num,sum;bool check(int line,int row,int k){int i,j;for(i=0;i<line;i++){if(x[i][row]==k) return false;}for(j=0;j<row;j++){if(x[line][j]==k) return false;}return true;}void Backtarck(int t){int i,j,k;if(t==num) {sum++;return;}i=t/m;j=t%m;  //行;列for(k=1;k<=n;k++){x[i][j]=k;if(check(i,j,k)) Backtarck(t+1);x[i][j]=0;}}int main(){while(~scanf("%d%d",&n,&m))  //m<n{sum=0;num=n*m;Backtarck(0);printf("%d\n",sum);}return 0;}

5-10 排列宝石问题

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int NM=25;int cor[NM][NM],size[NM][NM],vis[NM][NM];int num,sum,n;bool check(int line,int row,int c,int s){int i,j;for(i=0;i<line;i++){if(cor[i][row]==c || size[i][row]==s) return false;}for(j=0;j<row;j++){if(cor[line][j]==c || size[line][j]==s) return false;}return true;}void Backtarck(int t){int i,j,x,y;if(t==num){sum++;return;}x=t/n;y=t%n;for(i=1;i<=n;i++)for(j=1;j<=n;j++){if(!vis[i][j]){cor[x][y]=i;size[x][y]=j;vis[i][j]=1;if(check(x,y,i,j)) Backtarck(t+1);cor[x][y]=0;size[x][y]=0;vis[i][j]=0;}}}int main(){while(cin>>n) {num=n*n;sum=0;memset(vis,0,sizeof(vis));Backtarck(0);cout<<sum<<endl;}return 0;}

5-11 重复拉丁矩阵问题

理解一：第一行和第一列的宝石排列，不是按最小字典序，但还是最小到最大排列

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int NM=25;int x[NM][NM],sx[NM][NM],sy[NM][NM],c[NM],v[NM*NM],v1[NM*NM];int tt[NM],n,m,ans,num,sum;bool check(int line,int row,int k){int i,j;if(sx[line][k]>c[k] || sy[row][k]>c[k]) return false;/*解法一if(line==0){for(j=0;j<row;j++)if(x[0][j]>k) return false;}else if(row==0){for(i=0;i<line;i++)if(x[i][0]>k) return false;}else {if(line>0){for(j=0;j<m-1;j++)  //m-1if(x[0][j]>x[0][j]) return false;}if(row>0){for(i=0;i<line;i++)  //lineif(x[i][0]>x[i+1][0]) return false;}}*//*解法二*/if(line==0){for(i=0;i<row;i++){if(x[0][i]>k) return false;}}if(row==0){for(j=0;j<line;j++){if(x[j][0]>k) return false;}}return true;}void Backtarck(int t){int i,j,k;if(t==num) {for(i=0;i<num;i++) v1[i]=v[i];sum++;return;}i=t/m;j=t%m;  //行;列for(k=1;k<=ans;k++){x[i][j]=k;sx[i][k]++;sy[j][k]++;v[t]=k;if(check(i,j,k)) Backtarck(t+1);x[i][j]=0;sx[i][k]--;sy[j][k]--;}}int main(){int i,j,k;while(~scanf("%d%d%d",&n,&m,&ans))  //m<n{k=0;for(i=1;i<=ans;i++){scanf("%d",&c[i]);for(j=0;j<c[i];j++) tt[k++]=i;}sum=0;num=n*m;for(i=0;i<m;i++) x[0][i]=tt[i];for(i=0;i<n;i++) x[i][0]=tt[i];Backtarck(0);printf("%d\n",sum);for(i=0;i<num;i++){cout<<v1[i]<<" ";if((i+1)%m==0)cout<<endl;}}return 0;}

理解二：同书本

#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int NM=25;int x[NM][NM],sx[NM][NM],sy[NM][NM],c[NM],v[NM*NM],v1[NM*NM];int tt[NM],n,m,ans,num,sum;bool check(int line,int row,int k){if(sx[line][k]>c[k] || sy[row][k]>c[k]) return false;return true;}void Backtarck(int t){int i,j,k;if(t==num) {for(i=0;i<num;i++) v1[i]=v[i];sum++;return;}i=t/m;j=t%m;  //行;列if(x[i][j]) {v[t]=x[i][j];Backtarck(t+1);}elsefor(k=1;k<=ans;k++){x[i][j]=k;sx[i][k]++;sy[j][k]++;v[t]=k;if(check(i,j,k)) Backtarck(t+1);x[i][j]=0;sx[i][k]--;sy[j][k]--;}}int main(){int i,j,k;while(~scanf("%d%d%d",&n,&m,&ans))  //m<n{k=0;memset(sx,0,sizeof(sx));memset(sy,0,sizeof(sy));for(i=1;i<=ans;i++){scanf("%d",&c[i]);for(j=0;j<c[i];j++) tt[k++]=i;}sum=0;num=n*m;//先把第一行和第一列的宝石放好for(i=0;i<m;i++) {x[0][i]=tt[i];sx[0][tt[i]]=1;sy[i][tt[i]]=1;}for(i=0;i<n;i++) {x[i][0]=tt[i];sy[0][tt[i]]=1;sx[i][tt[i]]=1;}Backtarck(0);printf("%d\n",sum);for(i=0;i<num;i++){cout<<v1[i]<<" ";if((i+1)%m==0)cout<<endl;}}return 0;}/*4 7 32 2 31 2 21 12 3 31 1 12 4 4 1 1 1 1*/