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=;
 const int INF=0x3f3f3f3f;
 const int MOD=,STA=;
 const LL LNF=1LL<<;
 const double EPS=1e-;
 const double OO=1e15;
 const int dx[]={-,,,};
 const int dy[]={,,,-};
 const int day[]={,,,,,,,,,,,,};
 //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*)<<],t[N][],q[N],id[N*];
 int n;

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

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

 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)>>;
     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)>>,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=;
         for(i=;i<=n;i++){
             scanf("%d%d",&t[i][],&t[i][]);
             id[k++]=t[i][];
             id[k++]=t[i][];
         }
         scanf("%d",&m);
         for(i=;i<m;i++){
             scanf("%d",&q[i]);
             id[k++]=q[i];
         }
         sort(id,id+k);
         k=unique(id,id+k)-id;
         mem(c,-);
         for(i=;i<=n;i++){
             L=lower_bound(id,id+k,t[i][])-id+;
             R=lower_bound(id,id+k,t[i][])-id+;
             update(,k,,L,R,i);
         }

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