2004-2005 Summer Petrozavodsk Camp, Andrew Stankevich Contest 9 (ASC 9)

F Restore the Tree

一棵树有n个叶子，已知叶子间两两的距离，求这棵树的一个构造方案。
我的做法：
（已经被查出反例）
首先对于每个点向外延伸点，每次把距离最近的2个点合并
最后考察其余的距离是否合法

G Unrhymable Rhymes

每次遇到一对取出。

#include<bits/stdc++.h> using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define ForkD(i,k,n) for(int i=n;i>=k;i--)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  #define Lson (o<<1)#define Rson ((o<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,0x3f,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define MEMx(a,b) memset(a,b,sizeof(a));#define INF (0x3f3f3f3f)#define F (1000000007)#define pb push_back#define mp make_pair#define fi first#define se second#define vi vector<int> #define pi pair<int,int>#define SI(a) ((a).size())#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;#define PRi2D(a,n,m) For(i,n) { \                        For(j,m-1) cout<<a[i][j]<<' ';\                        cout<<a[i][m]<<endl; \                        } #pragma comment(linker, "/STACK:102400000,102400000")#define ALL(x) (x).begin(),(x).end()#define gmax(a,b) a=max(a,b);#define gmin(a,b) a=min(a,b);typedef long long ll;typedef long double ld;typedef unsigned long long ull;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return ((a-b)%F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}inline int read(){    int x=0,f=1; char ch=getchar();    while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}    while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}    return x*f;} int n;int a[10000],a2[10000];int c[10000]={};vector<int> v[5000];int tcase[10000]={};int main(){    freopen("rhymes.in","r",stdin);    freopen("rhymes.out","w",stdout);    int n=read();    For(i,n) a[i]=read(),a2[i]=a[i];    sort(a2+1,a2+1+n);    int m=unique(a2+1,a2+1+n)-a2-1;    For(i,n) a[i]=lower_bound(a2+1,a2+1+m,a[i])-a2;    int t=1,l=0;    vi tmp;    For(i,n) {        if (t!=tcase[a[i]]) tcase[a[i]]=t,c[a[i]]=0,v[a[i]].clear();        c[a[i]]++,v[a[i]].pb(i);        if (c[a[i]]%2==0) {            tmp.pb(v[a[i]][SI(v[a[i]])-2]);            tmp.pb(v[a[i]][SI(v[a[i]])-1]);            ++l;            if (l%2==0) ++t;        }       }    printf("%d\n",l/2);    int sz=SI(tmp);sort(ALL(tmp));    if (sz%4!=0) sz-=2;    Rep(i,sz) {        printf("%d",tmp[i]);        if (i%4==3) puts("");else putchar(' ');    }    return 0;}