[线段树+离散化]LightOJ 1089 - Points in Segments (II)
题目描述：
Givennsegments (1 dimensional) andqpoints, for each point you have to find the number of segments which contain that point. A pointpiwill lie in a segmentA BifA ≤ pi≤ B.
For example, if the segments are(6 12), (8 8), (10 12), (8 11), (0 12)and the point is11, then it is contained by4segments.
Input
Input starts with an integerT (≤ 5), denoting the number of test cases.
Each case starts with a line containing two integersn (1 ≤ n ≤ 50000)andq (1 ≤ q ≤ 50000).
Each of the nextnlines contains two integersAkBk(0 ≤ Ak≤ Bk≤ 108)denoting a segment.
Each of the nextqlines contains an integer denoting a point. Each of them range in[0, 108].
Output
For each case, print the case number in a single line. Then for each point, print the number of segments that contain that point.
Sample Input |
Output for Sample Input |
| 1 5 4 6 12 8 8 10 12 8 11 0 12 11 12 2 20 |
Case 1: 4 3 1 0 |
大意：先给定若干条线段，然后给定Q个询问，询问的形式为给定一个点，输出该点被线段覆盖多少次。
线段树功能：查询，成段更新
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const int MAXN = 222222;
int arr[MAXN];
struct Seg{
int a,b;
}s[MAXN];
int pt[MAXN],cov[MAXN<<2],N,T,S,Q;
void build(int l,int r,int rt){
cov[rt] = 0;
if(l==r)return;
int m = (l+r)>>1;
build(lson);
build(rson);
}
void pushDOWN(int rt){
if(cov[rt]){
cov[rt<<1]+=cov[rt];
cov[rt<<1|1]+=cov[rt];
cov[rt] = 0;
}
}
void update(int L,int R,int l,int r,int rt){
if(L<=l&&R>=r){
cov[rt]++;
return;
}
pushDOWN(rt);
int m = (l+r)>>1;
if(m>=L)update(L,R,lson);
if(m<R)update(L,R,rson);
}
int query(int pos,int l,int r,int rt){
if(l==r){
return cov[rt];
}
pushDOWN(rt);
int m = (l+r)>>1;
if(pos<=m)return query(pos,lson);
else return query(pos,rson);
}
int bs(int x){
int l = 0,r = N-1,m;
while(l<=r){
m = (l+r)>>1;
if(arr[m]==x)return m;
else if(arr[m]>x)r = m-1;
else l = m+1;
}
return -1;
}
int main(){
scanf("%d",&T);
for(int cas=1;cas<=T;cas++){
scanf("%d%d",&S,&Q);N = 0;
for(int i=0;i<S;i++){
scanf("%d%d",&s[i].a,&s[i].b);
arr[N++] = s[i].a;
arr[N++] = s[i].b;
}
for(int i=0;i<Q;i++){
scanf("%d",&pt[i]);
arr[N++] = pt[i];
}
sort(arr,arr+N);
int temp = 1;
for(int i=1;i<N;i++)if(arr[i]!=arr[i-1])arr[temp++] = arr[i];
N = temp;
build(0,N-1,1);
for(int i=0;i<S;i++){
update(bs(s[i].a),bs(s[i].b),0,N-1,1);
}
printf("Case %d:\n",cas);
for(int i=0;i<Q;i++){
printf("%d\n",query(bs(pt[i]),0,N-1,1));
}
}
return 0;
}