线段树+离散化poj2528

http://poj.org/problem?id=2528

题意是给你n个区间，后来的区间会覆盖前面的区间，问你最后有多少区间没有被完全覆盖

由于r的最大值是10000000，所以先将数据进行离散化处理，把所有所有区间的l,r排序，离散化后在从后往前去覆盖区间

如果该区间已经被覆盖，则返回false

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

struct Point
{
    int x;
    int id;
} point[4*20005];

struct Tree
{
    int l,r;
    int vis;
} tree[4*20005];
bool cmp1(Point a,Point b)
{
    return a.x<b.x;
}
bool cmp2(Point a,Point b)
{
        if(a.id==b.id)
        {
                return a.x<b.x;
        }
    return a.id>b.id;
}
void PushUp(int rt)
{
        tree[rt].vis=tree[2*rt].vis&&tree[2*rt+1].vis;
}
void bulid(int rt,int L,int R)
{
        tree[rt].vis=0;
        tree[rt].l=L;
        tree[rt].r=R;
        if(tree[rt].l==tree[rt].r)
        {
                return;
        }
        int mid=(L+R)/2;
        bulid(2*rt,L,mid);
        bulid(2*rt+1,mid+1,R);
}
bool query(int rt,int L,int R)
{
        if(tree[rt].vis)
        {
                return false;
        }
        if(tree[rt].l==L&&tree[rt].r==R)
        {
                tree[rt].vis=1;
                return true;
        }
        bool isok=false;
        int mid=(tree[rt].l+tree[rt].r)/2;
        if(R<=mid)
        {
                isok = query(2*rt,L,R);
        }
        else if(L>=mid+1)
        {
                isok = query(2*rt+1,L,R);
        }
        else
        {
                isok = query(2*rt,L,mid)|query(2*rt+1,mid+1,R);
        }
        PushUp(rt);
        return isok;
}
int main()
{

    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        for (int i=0; i<2*n ; i+=2 )
        {
            scanf("%d %d",&point[i].x,&point[i+1].x);
            point[i].id=point[i+1].id=i;
        }
        sort(point,point+2*n,cmp1);
        int pre=0,cnt=0;
        for (int i=0; i<2*n ; i++ )
        {
            if(point[i].x==pre)
            {
                point[i].x=cnt;
            }
            else
            {
                pre=point[i].x;
                point[i].x=++cnt;
            }
        }
        int ans=0;
        bulid(1,1,cnt);
        sort(point,point+2*n,cmp2);
        for (int i=0; i<2*n ; i+=2 )
        {
                if(query(1,point[i].x,point[i+1].x))
                {
                        ans++;
                }
        }
        printf("%d
",ans);
    }
    return 0;
}