天梯赛练习题L2（006-010）
L2-006 树的遍历
- 先序遍历：根→左→右
- 中序遍历：左→根→右
- 后序遍历：左→右→根
- 层序遍历：按照每层顺序遍历
- 根据后序遍历和中序遍历，可以知道：
- 1.后序遍历的最后一个值，一定是根节点；
- 2.此根节点在中序遍历中，将左右子树划分开来；
- 3.在左右子树中，对应的区间结点数一定是相同的，于是再重复1.2.的过程即可；
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18;
const LL N=500200,M=2002;
LL n,post[N],in[N];
map<LL,LL> L,R;
int dfs(int l1,int r1,int l2,int r2)
{
if(l1>r1) return 0;
int root=post[r1];
int mid;
for(mid=l2;mid<=r2&&in[mid]!=root;mid++) ;
L[root]=dfs(l1,l1-l2+mid-1,l2,mid-1);
R[root]=dfs(r1-r2+mid,r1-1,mid+1,r2);
return root;
}
void bfs(int root)
{
queue<int> q;
q.push(root);
bool flag=false;
while(!q.empty())
{
int top=q.front();
if(flag) cout<<" ";
cout<<top;
flag=true;
q.pop();
if(L[top]) q.push(L[top]);
if(R[top]) q.push(R[top]);
}
cout<<endl;
}
int main()
{
cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
LL T=1;
//cin>>T;
while(T--)
{
cin>>n;
for(int i=0;i<n;i++)//后序遍历
cin>>post[i];
for(int i=0;i<n;i++)//中序遍历
cin>>in[i];
bfs(dfs(0,n-1,0,n-1));
}
return 0;
}
L2-007 家庭房产
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18;
const LL N=500200,M=2002;
LL num[N],sum[N],father[N],vis[N];
struct node
{
LL minn,people,avgwu,avgsum;
double x1,x2;
const bool operator<(const node &A) const {
if(A.x2==x2) return minn<A.minn;
return x2>A.x2;
}
}a[N];
int find(int x)
{
if(father[x]!=x) father[x]=find(father[x]);
return father[x];
}
void merge(int x,int y)
{
LL fx=find(x),fy=find(y);
if(fx>fy) father[fx]=fy;
else father[fy]=fx;
}
int main()
{
//cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
LL T=1;
//cin>>T;
while(T--)
{
LL n;
cin>>n;
for(int i=0;i<=9999;i++)
father[i]=i;
while(n--)
{
LL idx,fa,ma,zi;
cin>>idx>>fa>>ma>>zi;
vis[idx]=1;
if(fa!=-1) merge(idx,fa),vis[fa]=1;
if(ma!=-1) merge(idx,ma),vis[ma]=1;
while(zi--)
{
LL x;
cin>>x;
merge(idx,x);
vis[x]=1;
}
cin>>num[idx]>>sum[idx];
}
for(int i=0;i<=9999;i++)
{
if(vis[i]==1)
{
LL fi=find(i);
a[fi].people++;
a[fi].avgwu+=num[i];
a[fi].avgsum+=sum[i];
}
}
vector<node> v;
for(int i=0;i<=9999;i++)
{
if(vis[i]&&find(i)==i)
{
a[i].minn=i;
a[i].x1=a[i].avgwu*1.0/a[i].people;
a[i].x2=a[i].avgsum*1.0/a[i].people;
v.push_back(a[i]);
}
}
sort(v.begin(),v.end());
cout<<v.size()<<endl;
for(int i=0;i<v.size();i++)
{
printf("%04d %d %.3lf %.3lf\n",v[i].minn,v[i].people,v[i].x1,v[i].x2);
}
}
return 0;
}
L2-008 最长对称子串
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18;
const LL N=500200,M=2002;
int main()
{
cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
LL T=1;
//cin>>T;
while(T--)
{
string s;
getline(cin,s);
LL maxn=1;
for(int i=0;i<s.size();i++)
{
LL l=i,r=i+1;
while(l>=0&&r<s.size()&&s[l]==s[r])
{
maxn=max(maxn,r-l+1);
l--,r++;
}
l=i-1,r=i+1;
while(l>=0&&r<s.size()&&s[l]==s[r])
{
maxn=max(maxn,r-l+1);
l--,r++;
}
}
cout<<maxn<<endl;
}
return 0;
}
L2-009 抢红包
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18;
const LL N=500200,M=2002;
LL n;
struct node
{
LL idx;
LL sum;
LL num;
}a[N];
bool cmp(node l,node r)
{
if(l.sum!=r.sum) return l.sum>r.sum;
else if(l.num!=r.num) return l.num>r.num;
else return l.idx<r.idx;
}
int main()
{
cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
LL T=1;
//cin>>T;
while(T--)
{
cin>>n;
for(int i=1;i<=n;i++)
{
a[i].idx=i;
}
for(int i=1;i<=n;i++)
{
LL x;
cin>>x;
while(x--)
{
LL who,money;
cin>>who>>money;
a[who].sum+=money;
a[who].num++;
a[i].sum-=money;
}
}
sort(a+1,a+1+n,cmp);
for(int i=1;i<=n;i++)
{
printf("%d %.2lf\n",a[i].idx,a[i].sum*1.0/100);
}
}
return 0;
}
L2-010 排座位
为什么要用并查集？因为朋友关系具有传递性
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,LL> PII;
const LL MAXN=1e18;
const LL N=500200,M=2002;
LL n,m,q,father[N];
LL g[M][M];
int find(int x)
{
if(father[x]!=x) father[x]=find(father[x]);
return father[x];
}
int main()
{
cin.tie(0); cout.tie(0); ios::sync_with_stdio(false);
LL T=1;
//cin>>T;
while(T--)
{
LL n,m,q;
cin>>n>>m>>q;
memset(g,0,sizeof g);
for(int i=1;i<=n;i++)
{
father[i]=i;
}
while(m--)
{
LL x,y,z;
cin>>x>>y>>z;
g[x][y]=g[y][x]=z;
if(z==1)
{
father[find(x)]=find(y);
}
}
while(q--)
{
LL x,y;
cin>>x>>y;
if(g[x][y]==1) cout<<"No problem"<<endl;
else if(g[x][y]==-1)
{
LL fx=find(x);
LL fy=find(y);
if(fx==fy) cout<<"OK but..."<<endl;
else cout<<"No way"<<endl;
}
else cout<<"OK"<<endl;
}
}
return 0;
}
