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CodeForces 19D Points（离散化+线段树+单点更新）
题目链接：
huangjing
题意：给了三种操作
1：add（x,y）将这个点增加二维坐标系
2：remove（x,y）将这个点从二维坐标系移除。
3：find（x，y）就是找到在（x，y）右上方的第一个点。
思路：我们能够建立n个set以x为横坐标，那么我们这个题就转化为找一个最小的x是否存在满足条件，那么x一旦被找到。那么纵坐标就自然而然的找到了。当然更新操作就是对maxy的维护，然后查询操作就是找出一个最小的x。。
还有由于n很大，所以要採用离散化的方法。然后进行离线处理。还是就是掌握set的使用方法，看来得买本c++primer了。insert（int val）将val插入到set中。且保证有序，erase（int val）将val删除且保证有序。。
题目：
Pete and Bob invented a new interesting game. Bob takes a sheet of paper and locates a Cartesian coordinate system on it as follows: point (0,?0) is located in the bottom-left corner, Ox axis is directed right, Oy axis is directed up. Pete gives Bob requests of three types:
- add x y — on the sheet of paper Bob marks a point with coordinates (x,?y). For each request of this type it‘s guaranteed that point(x,?y) is not yet marked on Bob‘s sheet at the time of the request.
- remove x y — on the sheet of paper Bob erases the previously marked point with coordinates (x,?y). For each request of this type it‘s guaranteed that point (x,?y) is already marked on Bob‘s sheet at the time of the request.
- find x y — on the sheet of paper Bob finds all the marked points, lying strictly above and strictly to the right of point (x,?y). Among these points Bob chooses the leftmost one, if it is not unique, he chooses the bottommost one, and gives its coordinates to Pete.
Bob managed to answer the requests, when they were 10, 100 or 1000, but when their amount grew up to 2·105, Bob failed to cope. Now he needs a program that will answer all Pete‘s requests. Help Bob, please!
The first input line contains number n (1?≤?n?≤?2·105) — amount of requests. Then there follow n lines — descriptions of the requests.add x y describes the request to add a point, remove x y — the request to erase a point, find x y — the request to find the bottom-left point. All the coordinates in the input file are non-negative and don‘t exceed 109.
For each request of type find x y output in a separate line the answer to it — coordinates of the bottommost among the leftmost marked points, lying strictly above and to the right of point (x,?y). If there are no points strictly above and to the right of point (x,?y), output -1.
7 add 1 1 add 3 4 find 0 0 remove 1 1 find 0 0 add 1 1 find 0 0
1 1 3 4 1 1
13 add 5 5 add 5 6 add 5 7 add 6 5 add 6 6 add 6 7 add 7 5 add 7 6 add 7 7 find 6 6 remove 7 7 find 6 6 find 4 4
7 7 -1 5 5
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
#include<cmath>
#include<string>
#include<queue>
#define eps 1e-9
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;
const int maxn=200000+10;
struct Tree
{
int x,y;
char op[5];
void init()
{
scanf("%s%d%d",op,&x,&y);
}
}tree[maxn<<2];
set<int>xx[maxn];
int maxy[maxn<<2],top[maxn],n,m;
void buildtree(int l,int r,int dex)
{
maxy[dex]=-1;
if(l==r) return;
int mid=(l+r)>>1;
buildtree(l,mid,dex<<1);
buildtree(mid+1,r,dex<<1|1);
}
void update(int l,int r,int dex,int pos)
{
if(l==r)
{
if(xx[pos].size())
maxy[dex]=*(--xx[pos].end());
else
maxy[dex]=-1;
return;
}
int mid=(l+r)>>1;
if(pos<=mid) update(l,mid,dex<<1,pos);
else update(mid+1,r,dex<<1|1,pos);
maxy[dex]=max(maxy[dex<<1],maxy[dex<<1|1]);
}
int Query(int l,int r,int dex,int L,int R,int val)
{
if(maxy[dex]<=val||L>R) return -1;//这里L>R是一个trick，wa在了第4组数据，有可能L>R
if(l==r) return l;
int mid=(l+r)>>1;
if(R<=mid) return Query(l,mid,dex<<1,L,R,val);
else if(L>mid) return Query(mid+1,r,dex<<1|1,L,R,val);
else
{
int t=Query(l,mid,dex<<1,L,R,val);
if(t!=-1) return t;
return Query(mid+1,r,dex<<1|1,L,R,val);
}
}
int main()
{
while(~scanf("%d",&n))
{
for(int i=1;i<=n;i++)
{
tree[i].init();
top[i]=tree[i].x;
}//离散化
sort(top+1,top+1+n);
m=unique(top+1,top+1+n)-(top+1);
for(int i=1;i<=m;i++)
xx[i].clear();
buildtree(1,m,1);
for(int i=1;i<=n;i++)
{
int pos=lower_bound(top+1,top+1+m,tree[i].x)-top;
if(tree[i].op[0]==‘a‘)
{
xx[pos].insert(tree[i].y);
update(1,m,1,pos);
}
else if(tree[i].op[0]==‘r‘)
{
xx[pos].erase(tree[i].y);
update(1,m,1,pos);
}
else
{
int ans=Query(1,m,1,pos+1,m,tree[i].y);
if(ans==-1)
printf("-1\n");
else
printf("%d %d\n",top[ans],*xx[ans].upper_bound(tree[i].y));
}
}
}
return 0;
}
CodeForces 19D Points（离散化+线段树+单点更新）