URAL-1987 Nested Segments 线段树简单区间覆盖

　　题目链接：http://acm.timus.ru/problem.aspx?space=1&num=1987

　　题意：给定n条线段，每两条线段要么满足没有公共部分，要么包含。给出m个询问，求当前点被覆盖的最小长度的线段编号。

　　由于线段不存在部分相交的情况，因此，直接按照输入顺序覆盖区间就可以了，因为后覆盖的线段更短。

//STATUS:C++_AC_187MS_6805KB
#include <functional>
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,102400000")
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef __int64 LL;
typedef unsigned __int64 ULL;
//const
const int N=100010;
const int INF=0x3f3f3f3f;
const int MOD=95041567,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End

int c[(N*3)<<2],t[N][2],q[N],id[N*3];
int n;

void pushdown(int rt)
{
    if(c[rt]!=-1)
        c[rt<<1]=c[rt<<1|1]=c[rt];
}

void pushup(int rt)
{
    if(c[rt<<1]==c[rt<<1|1])
        c[rt]=c[rt<<1];
    else c[rt]=-1;
}

void update(int l,int r,int rt,int L,int R,int val)
{
    if(L<=l && r<=R){
        c[rt]=val;
        return;
    }
    pushdown(rt);
    int mid=(l+r)>>1;
    if(L<=mid)update(lson,L,R,val);
    if(R>mid)update(rson,L,R,val);
    pushup(rt);
}

int query(int l,int r,int rt,int w)
{
    if(l==r){
        return c[rt];
    }
    pushdown(rt);
    int mid=(l+r)>>1,ret;
    if(w<=mid)ret=query(lson,w);
    else ret=query(rson,w);
    pushup(rt);
    return ret;
}

int main()
{
 //   freopen("in.txt","r",stdin);
    int i,j,k,L,R,m;
    while(~scanf("%d",&n))
    {
        k=0;
        for(i=1;i<=n;i++){
            scanf("%d%d",&t[i][0],&t[i][1]);
            id[k++]=t[i][0];
            id[k++]=t[i][1];
        }
        scanf("%d",&m);
        for(i=0;i<m;i++){
            scanf("%d",&q[i]);
            id[k++]=q[i];
        }
        sort(id,id+k);
        k=unique(id,id+k)-id;
        mem(c,-1);
        for(i=1;i<=n;i++){
            L=lower_bound(id,id+k,t[i][0])-id+1;
            R=lower_bound(id,id+k,t[i][1])-id+1;
            update(1,k,1,L,R,i);
        }

        for(i=0;i<m;i++){
            printf("%d
",query(1,k,1,lower_bound(id,id+k,q[i])-id+1));
        }
    }
    return 0;
}