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A.BZOJ 3033: 太鼓达人
题目
代码
#include<bits/stdc++.h>
using namespace std;
const int maxn=2e3+50;
template<typename T>inline void read(T &x)
{
x=0;
T f=1,ch=getchar();
while (!isdigit(ch) && ch^'-') ch=getchar();
if (ch=='-') f=-1, ch=getchar();
while (isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48), ch=getchar();
x*=f;
}
int n,t,ans[maxn];
bool used[maxn];
inline bool eular(int x,int k)//x是当前子串的十进制表示，k是当前查询次
{
if (used[x]) return 0;
used[x]=1,ans[k]=x&1;//统计答案，&的规则是：有零则零，否则为一,返回末位数字
if (k==t) return 1;//查询结束
if (eular((x<<1)&(t-1),k+1)) return 1;
if (eular((x<<1|1)&(t-1),k+1)) return 1;//将每个二进制数看成一个点，将他前面的数删去，并在它后面加上0/1就能得到两个新数
used[x]=0;//回溯，以免影响之后的dfs
return 0;
}
int main()
{
read(n);
t=1<<n;
printf("%d ",t);
eular(t-2,1);//从每位走n位子串，可得到的完整子串数量为2^n-2
for (int i=1;i<=t;++i)
printf("%d",ans[i]);
return 0;
}
邱神版本：
#include<bits/stdc++.h>
using namespace std;
int read()
{
int a;
cin>>a;
return a;
}
int i,k,m;
struct node
{
int x,deep;
}o[10010];
bool Orz(node a,node b)
{
return a.deep<b.deep;
}
void dfs(int now,int deep)
{
if(deep==m+1)
{
now=now%(m/2);
now=now*2;
if(now)
return ;
sort(o,o+m,Orz);
for(i=1;i<k;i++)
cout<<0;
for(i=0;i<=m-k;i++)
cout<<o[i].x%2;
exit(0) ;
}
//这里应该有一个关于deep和now的if使得dfs停下来并输出
now=now%(m/2);
now=now*2;
if(!o[now].deep)
{
o[now].deep=deep;
dfs(now,deep+1);
o[now].deep=0;
}
now++;
if(!o[now].deep)
{
o[now].deep=deep;
dfs(now,deep+1);
o[now].deep=0;
}
return ;
}
int main()
{
//freopen("123.in","r",stdin);
k=read();m=int(pow(2,k*1.0));
cout<<m<<' ';
for(i=0;i<m;i++)
o[i].x=i;
o[0].deep=1;
dfs(0,2);
}
B. POJ 1386 Play on Words
题目
题解
其实就是让判断是否是欧拉回路或欧拉通路.
建图需要一点思维，把26个字母当成是节点，每个单词当成是一条有向边。
比如单词
就是一条
到
的有向边。
然后用并查集判断图是否联通即可。
代码
#include<algorithm>
#include<bitset>
#include<cctype>
#include<cerrno>
#include<clocale>
#include<cmath>
#include<complex>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#include<deque>
#include<exception>
#include<fstream>
#include<functional>
#include<limits>
#include<list>
#include<map>
#include<iomanip>
#include<ios>
#include<iosfwd>
#include<iostream>
#include<istream>
#include<ostream>
#include<queue>
#include<set>
#include<sstream>
#include<stack>
#include<stdexcept>
#include<streambuf>
#include<string>
#include<utility>
#include<vector>
#include<cwchar>
#include<cwctype>
using namespace std;
const int maxn=30;
template<typename T>inline void read(T &x)
{
x=0;
T f=1,ch=getchar();
while (!isdigit(ch) && ch^'-') ch=getchar();
if (ch=='-') f=-1, ch=getchar();
while (isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48), ch=getchar();
x*=f;
}
int fa[maxn];
inline int get(int x)
{
if (x==fa[x]) return x;
return fa[x]=get(fa[x]);
}
int indeg[maxn],outdeg[maxn];
int main()
{
int t;read(t);
while (t--)
{
int n;read(n);
memset(indeg,0,sizeof(indeg));
memset(outdeg,0,sizeof(outdeg));
for (int i=0;i<26;++i)
fa[i]=i;
char str[1005];
for (int i=1;i<=n;++i)
{
scanf("%s",str);
int x=str[0]-'a';
int y=str[strlen(str)-1]-'a';
if (get(x)!=get(y))
fa[get(x)]=get(y);
++indeg[y],++outdeg[x];
}
bool flag=true;
int cnt=0,In=0,Out=0;
for (int i=0;i<26;++i)
{
if (get(i)==i && (indeg[i]+outdeg[i])>0)
++cnt;
if (indeg[i]!=outdeg[i])
{
if (indeg[i]-1==outdeg[i])
++In;
else if (indeg[i]+1==outdeg[i])
++Out;
else
flag=false;
}
}
if (flag && cnt==1 && (In==1 && Out==1 || In==0 && Out==0))
puts("Ordering is possible.");//欧拉回路和通路都符合题目条件
else
puts("The door cannot be opened.");
}
return 0;
}
C. CF 516B. Drazil and Tiles
题目
代码
#include<algorithm>
#include<bitset>
#include<cctype>
#include<cerrno>
#include<clocale>
#include<cmath>
#include<complex>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#include<deque>
#include<exception>
#include<fstream>
#include<functional>
#include<limits>
#include<list>
#include<map>
#include<iomanip>
#include<ios>
#include<iosfwd>
#include<iostream>
#include<istream>
#include<ostream>
#include<queue>
#include<set>
#include<sstream>
#include<stack>
#include<stdexcept>
#include<streambuf>
#include<string>
#include<utility>
#include<vector>
#include<cwchar>
#include<cwctype>
using namespace std;
const int maxn=2e3+2;
const int dx[]={0,0,1,-1};
const int dy[]={1,-1,0,0};
const char ch[]="><v^";
const char str[]="<>^v";
template<typename T>inline void read(T &x)
{
x=0;
T f=1,ch=getchar();
while (!isdigit(ch) && ch^'-') ch=getchar();
if (ch=='-') f=-1, ch=getchar();
while (isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48), ch=getchar();
x*=f;
}
struct rec
{
int x,y;
}a[maxn*maxn];
int m,n,tot,G[maxn][maxn];
char c[maxn][maxn];
inline bool check(int x,int y)
{
return x>0 && x<=m && y>0 && y<=n;
}
inline void dfs(int x,int y)
{
int cnt=0;
for (int i=0;i<4;++i)
{
int xx=x+dx[i],yy=y+dy[i];
if (check(xx,yy) && c[xx][yy]=='.') ++cnt;
}
G[x][y]=cnt;
}
inline bool topsort()
{
queue<pair<int,int> >q;
int cnt=0;
for (int i=0;i<tot;++i)
{
int x=a[i].x,y=a[i].y;
if (G[x][y]==1)
{
q.push(pair<int,int> (x,y));
G[x][y]=0;
}
}
pair<int,int>m;
while (!q.empty())
{
m=q.front();
q.pop();
int x=m.first,y=m.second;
--G[x][y];
for (int i=0;i<4;++i)
{
int xx=x+dx[i],yy=y+dy[i];
if (check(xx,yy) && c[xx][yy]=='.')
{
G[xx][yy]=0;
c[xx][yy]=ch[i];
c[x][y]=str[i];
for (int j=0;j<4;++j)
{
int xxx=xx+dx[j],yyy=yy+dy[j];
if (check(xxx,yyy) && c[xxx][yyy]=='.')
{
dfs(xxx,yyy);
if (G[xxx][yyy]==1)
q.push(pair<int,int> (xxx,yyy));
}
}
cnt+=2;
}
}
}
return cnt==tot;
}
int main()
{
read(m);read(n);tot=0;
for (int i=1;i<=m;++i)
{
scanf("%s",c[i]+1);
for (int j=1;j<=n;++j)
if (c[i][j]=='.')
{
a[tot].x=i,a[tot++].y=j;
G[i][j]=0;
if (check(i-1,j) && c[i-1][j]=='.')
++G[i-1][j],++G[i][j];
if (check(i,j-1) && c[i][j-1]=='.')
++G[i][j-1],++G[i][j];
}
}
if (topsort())
for (int i=1;i<=m;++i)
printf("%s\n",c[i]+1);
else
puts("Not unique");
return 0;
}
D. POJ 3169 Layout
题目
代码
#include<algorithm>
#include<bitset>
#include<cctype>
#include<cerrno>
#include<clocale>
#include<cmath>
#include<complex>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#include<deque>
#include<exception>
#include<fstream>
#include<functional>
#include<limits>
#include<list>
#include<map>
#include<iomanip>
#include<ios>
#include<iosfwd>
#include<iostream>
#include<istream>
#include<ostream>
#include<queue>
#include<set>
#include<sstream>
#include<stack>
#include<stdexcept>
#include<streambuf>
#include<string>
#include<utility>
#include<vector>
#include<cwchar>
#include<cwctype>
using namespace std;
const int maxn=1e6+10;
const int inf=0x3f3f3f3f;
inline int read()
{
int f=1,num=0;
char ch=getchar();
while (!isdigit(ch)) { if (ch=='-') f=-1; ch=getchar(); }
while (isdigit(ch)) num=(num<<1)+(num<<3)+(ch^48), ch=getchar();
return num*f;
}
int ver[maxn*2],edge[maxn*2],Next[maxn*2],head[maxn],len=0;
inline void add(int x,int y,int z)
{
ver[++len]=y,edge[len]=z,Next[len]=head[x],head[x]=len;
}
int dist[maxn],cnt[maxn];
bool v[maxn];
queue<int>q;
int spfa(int s,int n)
{
for (int i=0;i<maxn;++i) dist[i]=inf;
memset(cnt,0,sizeof(cnt));
memset(v,0,sizeof(v));
++cnt[s];
dist[s]=0;
v[s]=1;
q.push(s);
while (!q.empty())
{
int x=q.front();
q.pop();
v[x]=0;
for (int i=head[x];i;i=Next[i])
{
int y=ver[i],z=edge[i];
if (dist[y]>dist[x]+z)
{
dist[y]=dist[x]+z;
if (!v[y])
{
v[y]=1,q.push(y);
++cnt[y];
if (cnt[y]>n) return -1;//存在负环
}
}
}
}
if (dist[n]==inf) return -2;
return dist[n];
}
int main()
{
int n=read(),x=read(),y=read();
for (int i=1;i<n;++i)
add(i+1,i,0);
for (int i=1;i<=x;++i)
{
int a=read(),b=read(),d=read();
add(a,b,d);
}
for (int i=1;i<=y;++i)
{
int a=read(),b=read(),d=read();
add(b,a,-d);
}
printf("%d\n",spfa(1,n));
return 0;
}