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。

  1 //STATUS:C++_AC_187MS_6805KB
  2 #include <functional>
  3 #include <algorithm>
  4 #include <iostream>
  5 //#include <ext/rope>
  6 #include <fstream>
  7 #include <sstream>
  8 #include <iomanip>
  9 #include <numeric>
 10 #include <cstring>
 11 #include <cassert>
 12 #include <cstdio>
 13 #include <string>
 14 #include <vector>
 15 #include <bitset>
 16 #include <queue>
 17 #include <stack>
 18 #include <cmath>
 19 #include <ctime>
 20 #include <list>
 21 #include <set>
 22 #include <map>
 23 using namespace std;
 24 //#pragma comment(linker,"/STACK:102400000,102400000")
 25 //using namespace __gnu_cxx;
 26 //define
 27 #define pii pair<int,int>
 28 #define mem(a,b) memset(a,b,sizeof(a))
 29 #define lson l,mid,rt<<1
 30 #define rson mid+1,r,rt<<1|1
 31 #define PI acos(-1.0)
 32 //typedef
 33 typedef __int64 LL;
 34 typedef unsigned __int64 ULL;
 35 //const
 36 const int N=100010;
 37 const int INF=0x3f3f3f3f;
 38 const int MOD=95041567,STA=8000010;
 39 const LL LNF=1LL<<60;
 40 const double EPS=1e-8;
 41 const double OO=1e15;
 42 const int dx[4]={-1,0,1,0};
 43 const int dy[4]={0,1,0,-1};
 44 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
 45 //Daily Use ...
 46 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 47 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 48 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 49 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 50 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 51 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 52 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 53 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 54 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 55 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 56 //End
 57 
 58 int c[(N*3)<<2],t[N][2],q[N],id[N*3];
 59 int n;
 60 
 61 void pushdown(int rt)
 62 {
 63     if(c[rt]!=-1)
 64         c[rt<<1]=c[rt<<1|1]=c[rt];
 65 }
 66 
 67 void pushup(int rt)
 68 {
 69     if(c[rt<<1]==c[rt<<1|1])
 70         c[rt]=c[rt<<1];
 71     else c[rt]=-1;
 72 }
 73 
 74 void update(int l,int r,int rt,int L,int R,int val)
 75 {
 76     if(L<=l && r<=R){
 77         c[rt]=val;
 78         return;
 79     }
 80     pushdown(rt);
 81     int mid=(l+r)>>1;
 82     if(L<=mid)update(lson,L,R,val);
 83     if(R>mid)update(rson,L,R,val);
 84     pushup(rt);
 85 }
 86 
 87 int query(int l,int r,int rt,int w)
 88 {
 89     if(l==r){
 90         return c[rt];
 91     }
 92     pushdown(rt);
 93     int mid=(l+r)>>1,ret;
 94     if(w<=mid)ret=query(lson,w);
 95     else ret=query(rson,w);
 96     pushup(rt);
 97     return ret;
 98 }
 99 
100 int main()
101 {
102  //   freopen("in.txt","r",stdin);
103     int i,j,k,L,R,m;
104     while(~scanf("%d",&n))
105     {
106         k=0;
107         for(i=1;i<=n;i++){
108             scanf("%d%d",&t[i][0],&t[i][1]);
109             id[k++]=t[i][0];
110             id[k++]=t[i][1];
111         }
112         scanf("%d",&m);
113         for(i=0;i<m;i++){
114             scanf("%d",&q[i]);
115             id[k++]=q[i];
116         }
117         sort(id,id+k);
118         k=unique(id,id+k)-id;
119         mem(c,-1);
120         for(i=1;i<=n;i++){
121             L=lower_bound(id,id+k,t[i][0])-id+1;
122             R=lower_bound(id,id+k,t[i][1])-id+1;
123             update(1,k,1,L,R,i);
124         }
125 
126         for(i=0;i<m;i++){
127             printf("%d
",query(1,k,1,lower_bound(id,id+k,q[i])-id+1));
128         }
129     }
130     return 0;
131 }