POJ 2528 Mayor's posters （线段树+离散化）

题目链接：http://poj.org/problem?id=2528

题目大意： 

给出n个海报的左端，右端，按照顺序贴上海报

求贴完后可以看到多少海报？

方法：

先将海报数据离散化，然后线段树查询

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>

using namespace std;

const int maxn = 10010;
struct node
{
int l, r;
} Poster[maxn];
int Section[maxn*2];
int Hash[maxn*1000];
struct Tree
{
int l, r;
bool flag;
} ans[maxn*10];

void Build(int l, int r, int rt)
{
ans[rt].l = l, ans[rt].r = r;
ans[rt].flag = false;
if(l == r)
return;
int mid = (l + r) >> 1;
Build(l, mid, rt<<1);
Build(mid+1, r, rt<<1|1);
}

bool Judge(int l, int r, int rt)
{
if(ans[rt].flag)
return false;
if(l == ans[rt].l && r == ans[rt].r)
{
ans[rt].flag = true;
return true;
}
bool Flag;
int mid = (ans[rt].l + ans[rt].r) >> 1;
if(r <= mid)
Flag = Judge(l, r, rt<<1);
else if(l > mid)
Flag = Judge(l, r, rt<<1|1);
else
{
bool flag1 = Judge(l, mid, rt<<1);
bool flag2 = Judge(mid+1, r, rt<<1|1);
Flag = (flag1 || flag2);
}
if(ans[rt<<1].flag && ans[rt<<1|1].flag)
ans[rt].flag = true;
return Flag;
}

int main()
{
int T, n;
scanf("%d",&T);
while(T-- && scanf("%d",&n))
{
int Count = 0;
for(int i=0; i<n; i++)
{
scanf("%d%d",&Poster[i].l, &Poster[i].r);
Section[Count++] = Poster[i].l, Section[Count++] = Poster[i].r;
}

sort(Section, Section+Count);
Count = unique(Section, Section+Count) - Section;
for(int i=0; i<Count; i++)
Hash[Section[i]] = i;
// printf("%d\n",Count);
Build(0, Count-1, 1);
int ant = 0;
for(int i=n-1; i>=0; i--)
if(Judge(Hash[Poster[i].l], Hash[Poster[i].r], 1))
ant++;

printf("%d\n",ant);
}
return 0;
}