hdu Cows

Problem Description

Farmer John's cows have discovered that the clover growing along the ridge of the hill (which we can think of as a one-dimensional number line) in his field is particularly good. 

Farmer John has N cows (we number the cows from 1 to N). Each of Farmer John's N cows has a range of clover that she particularly likes (these ranges might overlap). The ranges are defined by a closed interval [S,E]. 

But some cows are strong and some are weak. Given two cows: cow_i and cow_j, their favourite clover range is [Si, Ei] and [Sj, Ej]. If Si <= Sj and Ej <= Ei and Ei - Si > Ej - Sj, we say that cow_i is stronger than cow_j. 

For each cow, how many cows are stronger than her? Farmer John needs your help!

 

Input

The input contains multiple test cases. 
For each test case, the first line is an integer N (1 <= N <= 10⁵), which is the number of cows. Then come N lines, the i-th of which contains two integers: S and E(0 <= S < E <= 10⁵) specifying the start end location respectively of a range preferred by some cow. Locations are given as distance from the start of the ridge. 

The end of the input contains a single 0.

 

Output

For each test case, output one line containing n space-separated integers, the i-th of which specifying the number of cows that are stronger than cow_i. 

 

Sample Input

31 20 33 40

 

Sample Output

1 0 0

 题目的意思是，求每个区间包含在几个区间里面....

#include<iostream>#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;#define maxx 100010int tree[maxx];struct Node{int x,y,i;}node[maxx];int lowbit(int v){return v&-v;}void add(int v,int var){while(v<=maxx){tree[v]+=var;v+=lowbit(v);}}int cmp(Node a,Node b){if(a.y==b.y)  return a.x<b.x;//这个要注意一下，如果相等，要把小的点放在前面 else return a.y>b.y;}int querry(int v){int sum=0;while(v>0){sum+=tree[v];v-=lowbit(v);}return sum;}int main(){int t,n,i,j,a,b,c[maxx];while(scanf("%d",&n),n){memset(tree,0,sizeof(tree));//初始化 memset(c,0,sizeof(c));for(i=1;i<=n;i++){node[i].i=i;//存位置，以便sort过后查找 scanf("%d%d",&node[i].x,&node[i].y);}sort(node+1,node+n+1,cmp);c[node[1].i]=querry(node[1].x+1);add(node[1].x+1,1);for(i=2;i<=n;i++){if(node[i].x==node[i-1].x && node[i].y==node[i-1].y){//判断重点 c[node[i].i]=c[node[i-1].i]; add(node[i].x+1,1);continue;}c[node[i].i]=querry(node[i].x+1);//计算前面的个数... add(node[i].x+1,1);}for(i=1;i<n;i++){printf("%d ",c[i]);}printf("%d\n",c[n]);}return 0;}