POJ2481：Cows(树状数组)

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 <= 10 ⁵), which is the number of cows. Then come N lines, the i-th of which contains two integers: S and E(0 <= S < E <= 10 ⁵) 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

Hint

Huge input and output,scanf and printf is recommended.

Source

POJ Contest,Author:Mathematica@ZSU

题意：

对于每头牛，吃草的区间在li,ri之间，问在n头牛中，对于第i头牛而言，有几头牛的区间大于这头牛

思路：

可以使用树状数组，先排好序，再统计其中一个即可

#include <iostream>#include <stdio.h>#include <string.h>#include <string>#include <stack>#include <queue>#include <map>#include <set>#include <vector>#include <math.h>#include <bitset>#include <list>#include <algorithm>#include <climits>using namespace std;#define lson 2*i#define rson 2*i+1#define LS l,mid,lson#define RS mid+1,r,rson#define UP(i,x,y) for(i=x;i<=y;i++)#define DOWN(i,x,y) for(i=x;i>=y;i--)#define MEM(a,x) memset(a,x,sizeof(a))#define W(a) while(a)#define gcd(a,b) __gcd(a,b)#define LL long long#define N 200005#define INF 0x3f3f3f3f#define EXP 1e-8#define lowbit(x) (x&-x)const int mod = 1e9+7;struct node{    int l,r,id;} a[N];int n,maxn,c[N];int cmp(node a,node b){    if(a.r!=b.r)        return a.r>b.r;    return a.l<b.l;}int sum(int x){    int ret = 0;    while(x>0)    {        ret+=c[x];        x-=lowbit(x);    }    return ret;}void add(int x,int d){    while(x<=maxn+1)    {        c[x]+=d;        x+=lowbit(x);    }}int ans[N];int main(){    int i,j,k,l,r;    while(~scanf("%d",&n),n)    {        MEM(ans,0);        MEM(c,0);        maxn = -1;        for(i = 1; i<=n; i++)        {            scanf("%d%d",&a[i].l,&a[i].r);            a[i].id = i;            maxn = max(maxn,a[i].r);        }        sort(a+1,a+1+n,cmp);        for(i = 1; i<=n; i++)        {            if(a[i].l == a[i-1].l && a[i].r == a[i-1].r)            {                ans[a[i].id] = ans[a[i-1].id];            }            else            {                ans[a[i].id] = sum(a[i].l+1);            }            add(a[i].l+1,1);        }        printf("%d",ans[1]);        for(i = 2; i<=n; i++)            printf(" %d",ans[i]);        printf("\n");    }    return 0;}