hdoj4325Flowers【线段树+离散化】

Flowers

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2603    Accepted Submission(s): 1281

Problem Description

As is known to all, the blooming time and duration varies between different kinds of flowers. Now there is a garden planted full of flowers. The gardener wants to know how many flowers will bloom in the garden in a specific time. But there are too many flowers in the garden, so he wants you to help him.

 

Input

The first line contains a single integer t (1 <= t <= 10), the number of test cases.
For each case, the first line contains two integer N and M, where N (1 <= N <= 10^5) is the number of flowers, and M (1 <= M <= 10^5) is the query times. 
In the next N lines, each line contains two integer S_i and T_i (1 <= S_i <= T_i <= 10^9), means i-th flower will be blooming at time [S_i, T_i].
In the next M lines, each line contains an integer T_i, means the time of i-th query.

 

Output

For each case, output the case number as shown and then print M lines. Each line contains an integer, meaning the number of blooming flowers.
Sample outputs are available for more details.

 

Sample Input

21 15 1042 31 44 8146

 

Sample Output

Case #1:0Case #2:121

 

Author

BJTU

 

Source

2012 Multi-University Training Contest 3

 

#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1#define inf 0x3f3f3f3fusing namespace std;const int maxn=1000010;int q[maxn];int sum[maxn<<2];int num[maxn<<1];struct Node{int s,e;}A[100010];int ans;bool cmp(int a,int b){return a<b;}int BS(int k,int n){int l=1,r=n;while(l<=r){int mid=(l+r)>>1;if(k==num[mid])return mid;else if(k<num[mid])r=mid-1;else l=mid+1;}}void update(int left,int right,int l,int r,int rt){if(left<=l&&r<=right){sum[rt]++;return ;}int mid=(l+r)>>1;if(left<=mid)update(left,right,lson);if(mid<right)update(left,right,rson);}void query(int pos,int l,int r,int rt){ans+=sum[rt];if(l==r)return;int mid=(l+r)>>1;if(pos<=mid)return query(pos,lson);else return query(pos,rson);}int main(){int n,m,i,j,k,t,test=1;scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);int cnt=1;for(i=0;i<n;++i){scanf("%d%d",&A[i].s,&A[i].e);num[cnt++]=A[i].s;num[cnt++]=A[i].e;}for(i=0;i<m;++i){scanf("%d",&q[i]);num[cnt++]=q[i];}sort(num,num+cnt,cmp);k=1;for(i=2;i<cnt;++i){if(num[i]!=num[i-1])num[++k]=num[i];}memset(sum,0,sizeof(sum));for(i=0;i<n;++i){int l=BS(A[i].s,k);int r=BS(A[i].e,k);update(l,r,1,k,1);}printf("Case #%d:\n",test++);for(i=0;i<m;++i){int pos=BS(q[i],k);ans=0;query(pos,1,k,1);printf("%d\n",ans);}}return 0;}