BZOJ 1901 Zju2112 Dynamic Rankings ——整体二分

【题目分析】

    上次用树状数组套主席树做的，这次用整体二分去水。

    把所有的查询的结果一起进行二分，思路很好。

【代码】

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>

#include <set>
#include <map>
#include <string>
#include <algorithm>
#include <vector>
#include <iostream>
#include <queue>
using namespace std;

#define maxn 100005
#define inf (0x3f3f3f3f)

int read()
{
int x=0,f=1; char ch=getchar();
while (ch<'0'||ch>'9') {if (ch=='-') f=-1; ch=getchar();}
while (ch>='0'&&ch<='9') {x=x*10+ch-'0'; ch=getchar();}
return x*f;
}

struct node{
int x,y,k;
int id,opt;
void print() {
printf("%d %d %d %d %d\n", x, y, k, id, opt);
}
}now;

int n,m,a[maxn],ans[maxn];
char s[15];
node q[maxn<<1],q1[maxn],q2[maxn];
int cnt=0,bit[maxn<<1],tot=0;

void add(int x,int f)
{
//  printf("add %d %d\n",x,f);
for (;x<=n;x+=x&-x) bit[x]+=f;
}

int sum(int x)
{
int ret=0;
//  printf("sum %d ",x);
for (;x;x-=x&-x) ret+=bit[x];
//  printf("is %d\n",x,ret);
return ret;
}

void solve(int ql,int qr,int l,int r)
{
//  printf("solve %d %d %d %d\n",ql,qr,l,r);
if (ql>qr) return ;
if (l==r)
{
for (int i=ql;i<=qr;++i)
if (q[i].opt==2) ans[q[i].id]=l;
return;
}
int mid=l+r>>1,p1=0,p2=0;
for (int i=ql;i<=qr;++i)
{
if (q[i].opt==1)
{
if (q[i].x<=mid)
{
add(q[i].id,q[i].y);
q1[++p1]=q[i];
}
else q2[++p2]=q[i];
}
else
{
int tmp=sum(q[i].y)-sum(q[i].x-1);
//          printf("tmp is %d\n");
if (q[i].k<=tmp) q1[++p1]=q[i];
else
{
//              cout<<q[i].x<<" "<<q[i].y<<"goto r "<<endl;
q[i].k-=tmp;
q2[++p2]=q[i];
}
}
}
for (int i=1;i<=p1;++i)
if (q1[i].opt==1)
add(q1[i].id,-q1[i].y);
for (int i=1;i<=p1;++i) q[ql+i-1]=q1[i];
for (int i=1;i<=p2;++i) q[ql+p1+i-1]=q2[i];
//  getchar();
solve(ql,ql+p1-1,l,mid);
solve(ql+p1,qr,mid+1,r);
}

int main()
{
n=read();m=read();
for (int i=1;i<=n;++i)
{
a[i]=now.x=read();now.y=1;now.k=inf;
now.id=i; now.opt=1;
q[++cnt]=now;
}
for (int i=1;i<=m;++i)
{
scanf("%s",s);
if (s[0]=='Q')
{
now.x=read();now.y=read();now.k=read();
now.id=++tot;now.opt=2;
q[++cnt]=now;
}
else
{
int xx=read();
now.x=a[xx];now.y=-1;now.k=inf;now.id=xx;now.opt=1;
q[++cnt]=now;
a[xx]=read();
now.x=a[xx];now.y=1;now.k=inf;now.id=xx;now.opt=1;
q[++cnt]=now;
}
}
//  for (int i=1;i<=cnt;++i) q[i].print();
solve(1,cnt,0,inf);
for (int i=1;i<=tot;++i) printf("%d\n",ans[i]);
}