标签：body eof ios lang fine idt i+1 call nal
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SuperMemo
Description Your friend, Jackson is invited to a TV show called SuperMemo in which the participant is told to play a memorizing game. At first, the host tells the participant a sequence of numbers, {A1, A2, ... An}. Then the host performs a series of operations and queries on the sequence which consists:
To make the show more interesting, the participant is granted a chance to turn to someone else that means when Jackson feels difficult in answering a query he may call you for help. You task is to watch the TV show and write a program giving the correct answer to each query in order to assist Jackson whenever he calls. Input The first line contains n (n ≤ 100000). The following n lines describe the sequence. Then follows M (M ≤ 100000), the numbers of operations and queries. The following M lines describe the operations and queries. Output For each "MIN" query, output the correct answer. Sample Input 5 1 2 3 4 5 2 ADD 2 4 1 MIN 4 5 Sample Output 5 Source
POJ Founder Monthly Contest – 2008.04.13, Yao Jinyu
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Splay裸题
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
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 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 (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (200000+10)
#define MAXM (100000+10)
typedef long long ll;
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+(a-b)/F*F+F)%F;}
int modF(int a,int b){return (a+a/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int n,m;
int a[MAXN];
class Splay
{
public:
int father[MAXN],siz[MAXN],n;
int ch[MAXN][2],val[MAXN];
bool root[MAXN],rev[MAXN];
int addv[MAXN],minv[MAXN];
int roo; //root
void mem(int _n)
{
MEM(father) MEM(siz) MEM(root) MEM(rev) MEM(ch) MEM(val) flag=0; MEM(addv) MEM(minv)
n=0;
roo=1;
build(roo,1,_n,0);root[1]=1;
}
void newnode(int &x,int f,int v)
{
x=++n;
father[x]=f;
val[x]=minv[x]=v;siz[x]=1;
}
void build(int &x,int L,int R,int f)
{
if (L>R) return ;
int m=(L+R)>>1;
newnode(x,f,a[m]);
build(ch[x][0],L,m-1,x);
build(ch[x][1],m+1,R,x);
maintain(x);
}
int getkth(int x,int k)
{
pushdown(x);
int t;
if (ch[x][0]) t=siz[ch[x][0]]; else t=0;
if (t==k-1) return x;
else if (t>=k) return getkth(ch[x][0],k);
else return getkth(ch[x][1],k-t-1);
}
void pushdown(int x)
{
if (x) if (rev[x])
{
swap(ch[x][0],ch[x][1]);
if (ch[x][0]) rev[ ch[x][0] ]^=1;
if (ch[x][1]) rev[ ch[x][1] ]^=1;
rev[x]^=1;
}
if (addv[x])
{
if (ch[x][0]) addv[ ch[x][0] ]+=addv[x],minv[ ch[x][0] ]+=addv[x],val[ ch[x][0] ]+=addv[x];
if (ch[x][1]) addv[ ch[x][1] ]+=addv[x],minv[ ch[x][1] ]+=addv[x],val[ ch[x][1] ]+=addv[x];
addv[x]=0;
}
}
void maintain(int x)
{
siz[x]=siz[ch[x][0]]+siz[ch[x][1]]+1;
minv[x]=val[x];
if (ch[x][0]) minv[ x ]=min(minv[x],minv[ ch[x][0] ] + addv[x] );
if (ch[x][1]) minv[ x ]=min(minv[x],minv[ ch[x][1] ] + addv[x] );
}
void rotate(int x)
{
int y=father[x],kind=ch[y][1]==x;
pushdown(y); pushdown(x);
ch[y][kind]=ch[x][!kind];
if (ch[y][kind]) {
father[ch[y][kind]]=y;
}
father[x]=father[y];
father[y]=x;
ch[x][!kind]=y;
if (root[y])
{
root[x]=1;root[y]=0;roo=x;
}
else
{
ch[father[x]][ ch[father[x]][1]==y ] = x;
}
maintain(y);maintain(x);
}
void splay(int x)
{
while(!root[x])
{
int y=father[x];
int z=father[y];
if (root[y]) rotate(x);
else if ( (ch[y][1]==x)^(ch[z][1]==y) )
{
rotate(x); rotate(x);
}
else
{
rotate(y); rotate(x);
}
}
roo=x;
}
void splay(int x,int r)
{
while(!(father[x]==r))
{
int y=father[x];
int z=father[y];
if (father[y]==r) rotate(x);
else if ( (ch[y][1]==x)^(ch[z][1]==y) )
{
rotate(x); rotate(x);
}
else
{
rotate(y); rotate(x);
}
}
}
void Cut(int a,int b,int c)
{
int x=getkth(roo,a),y=getkth(roo,b);
splay(x);
splay(y,roo);
pushdown(x);pushdown(y);
int z=ch[y][0];
ch[y][0]=0; maintain(y); maintain(x);
int u=getkth(roo,c),v=getkth(roo,c+1);
splay(u);
splay(v,roo);
pushdown(u);pushdown(v);
ch[v][0]=z;father[z]=v;
maintain(v);maintain(u);
}
void Flip(int a,int b)
{
int x=getkth(roo,a),y=getkth(roo,b);
splay(x);
splay(y,roo);
pushdown(x);pushdown(y);
int z=ch[y][0];
rev[z]^=1;
maintain(y); maintain(x);
}
void Add(int a,int b,int c)
{
int x=getkth(roo,a),y=getkth(roo,b);
splay(x);
splay(y,roo);
pushdown(x);pushdown(y);
int z=ch[y][0];
addv[z]+=c; val[z]+=c; minv[z]+=c;
maintain(y); maintain(x);
}
int queryMin(int a,int b)
{
int x=getkth(roo,a),y=getkth(roo,b);
splay(x);
splay(y,roo);
pushdown(x);pushdown(y);
int z=ch[y][0];
maintain(y); maintain(x);
return minv[z];
}
void insert(int a,int P)
{
int x=getkth(roo,a),y=getkth(roo,a+1);
splay(x);
splay(y,roo);
pushdown(x);pushdown(y);
newnode(ch[y][0],y,P);
maintain(y); maintain(x);
}
void Delete(int a,int b)
{
int x=getkth(roo,a),y=getkth(roo,b);
splay(x);
splay(y,roo);
pushdown(x);pushdown(y);
int z=ch[y][0];
ch[y][0]=0; father[z]=0; maintain(y); maintain(x);
}
bool flag;
void print(int x)
{
if (x==0) return ;
pushdown(x);
print(ch[x][0]);
printf("%d ",val[x]);
print(ch[x][1]);
}
}S;
char s[20];
int main()
{
// freopen("poj3580.in","r",stdin);
// freopen(".out","w",stdout);
while(cin>>n)
{
For(i,n) scanf("%d",&a[i+1]); a[1]=a[n+2]=INF;
S.mem(n+2);
cin>>m;
For(i,m)
{
scanf("%s",s);
if (s[0]==‘A‘) //ADD
{
int x,y,D;
scanf("%d%d%d",&x,&y,&D);
S.Add(x,y+2,D);
} else if (s[0]==‘I‘) { //INSERT
int x,P;
scanf("%d%d",&x,&P);
S.insert(x+1,P);
} else if (s[0]==‘D‘) { //DELETE
int x;
scanf("%d",&x);
S.Delete(x,x+2);
} else if (s[0]==‘M‘) { //MIN
int x,y;
scanf("%d%d",&x,&y);
printf("%d\n",S.queryMin(x,y+2));
} else if (s[3]==‘E‘) { //REVERSE
int x,y;
scanf("%d%d",&x,&y);
S.Flip(x,y+2);
} else { //REVOLVE
int x,y,t;
scanf("%d%d%d",&x,&y,&t);
t=(t%(y-x+1)+(y-x+1))%(y-x+1);
if (t) S.Cut(y+2-t-1,y+2,x);
}
// S.print(S.roo);cout<<endl;
}
}
return 0;
}
POJ 3580(SuperMemo-Splay区间加)[template:Splay V2]
标签：body eof ios lang fine idt i+1 call nal
原文地址：http://www.cnblogs.com/jhcelue/p/6846233.html
