算分课笔记
全排列
#include<iostream>
#include <stdio.h>
using namespace std;
/*算法保证了每次把第i个元素换到开头之前,这时候的序列都是1,2,3...
因为每个perm函数都是里面换了后会换回来,当一个perm结束后它对数列的顺序相当于没改变,只是在过程中把排列输出来了。*/
int a[100];
int casee;
void perm(int k,int m)//实现从下标k到下标m的全排列,每实现一个,输出下标从0到m的元素。
{
if(k==m)
{
for(int i=0;i<=m;++i)
{
cout<<a[i]<<" ";
}
cout<<"\t\t"<<++casee<<endl;
}
else
{
for(int i=k;i<=m;++i)
{
swap(a[k],a[i]);
perm(k+1,m);
swap(a[k],a[i]);
}
}
}
int main()
{
int n;cin>>n;
for(int i=0;i<n;++i)a[i]=i+1,cout<<a[i]<<" ";
cout<<endl<<"start"<<endl;
perm(0,n-1);
cout<<endl<<endl;
for(int i=0;i<n;++i)cout<<a[i]<<" ";//数组元素的顺序保持
return 0;
}
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棋盘覆盖
#include<iostream>
#include <stdio.h>
using namespace std;
const int N=100;
int tile=0,board[N][N];
void chessBoard(int tr,int tc,int dr,int dc,int size)//tr,tc是当前棋盘的左上角坐标,dr dc是特殊棋子坐标,size棋盘边长
{
if(size==1)return;
int t=++tile,s=size/2;// t为L块顺序放置的编号, s半边长
if(dr<tr+s&&dc<tc+s)chessBoard(tr,tc,dr,dc,s);//左上
else board[tr+s-1][tc+s-1]=t,chessBoard(tr,tc,tr+s-1,tc+s-1,s);
if(dr<tr+s&&dc>=tc+s)chessBoard(tr,tc+s,dr,dc,s);//右上
else board[tr+s-1][tc+s]=t,chessBoard(tr,tc+s,tr+s-1,tc+s,s);
if(dr>=tr+s&&dc<tc+s)chessBoard(tr+s,tc,dr,dc,s);//左下
else board[tr+s][tc+s-1]=t,chessBoard(tr+s,tc,tr+s,tc+s-1,s);
if(dr>=tr+s&&dc>=tc+s)chessBoard(tr+s,tc+s,dr,dc,s);//右下
else board[tr+s][tc+s]=t,chessBoard(tr+s,tc+s,tr+s,tc+s,s);
}
int main()
{
int k;
cout<<"输入棋盘大小参数k,边长为2^k"<<endl;
cin>>k;
cout<<"输入特殊棋子行列坐标(从0开始)"<<endl;
int dr,dc;
cin>>dr>>dc;
chessBoard(0,0,dr,dc,1<<k);
cout<<"需要L块个数:"<<tile<<endl;
for(int i=0;i<(1<<k);++i)
{
for(int j=0;j<(1<<k);++j)cout<<board[i][j]<<"\t";
cout<<"\n\n\n";
}
return 0;
}
-----------------------------------------------------------------------------------------
最长公共子序列
样例:
axaxacbdscs
axabss
#include<iostream>
#include<string.h>
using namespace std;
const int N=500;
int c[N][N];
void lcslength(char x[],char y[],int ** b)
{
int m=strlen(x)-1,n=strlen(y)-1;
for(int i=0;i<=m;++i)c[i][0]=0;
for(int i=0;i<=n;++i)c[0][i]=0;
for(int i=1;i<=m;++i)
{
for(int j=1;j<=n;++j)
{
if(x[i]==y[j])
{
c[i][j]=c[i-1][j-1]+1;
b[i][j]=1;
}
else if(c[i-1][j]>=c[i][j-1])
{
c[i][j]=c[i-1][j];
b[i][j]=2;
}
else
{
c[i][j]=c[i][j-1];
b[i][j]=3;
}
}
}
}
void lcs(int i,int j,char x[],int ** b)
{
if(i==0||j==0)return ;
if(b[i][j]==1)
{
lcs(i-1,j-1,x,b);
cout<<x[i];
}
else if(b[i][j]==2)lcs(i-1,j,x,b);
else if(b[i][j]==3)lcs(i,j-1,x,b);
}
int main()
{
char x[N],y[N];
int **b=new int *[N]; for(int i=0;i<N;++i)b[i]=new int [N];
cout<<"请输入两个字符串"<<endl;
cin>>x>>y;
int m=strlen(x),n=strlen(y);
for(int i=m-1;i>=0;--i)x[i+1]=x[i];x[m+1]='\0';//让字符从下标1开始存放,便于处理
for(int i=n-1;i>=0;--i)y[i+1]=y[i];y[n+1]='\0';
cout<<"两个字符串的长度分别为"<<m<<" "<<n<<endl;
lcslength(x,y,b);
cout<<"最长公共子序列的长度为"<<c[m][n]<<endl;
if(c[m][n]!=0) cout<<"最长公共子序列:";
lcs(m,n,x,b);
return 0;
}