2018.07.08 POJ 2481 Cows

Cows
Time Limit: 3000MS Memory Limit: 65536K
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: cowi and cowj, their favourite clover range is [Si, Ei] and [Sj, Ej]. If Si <= Sj and Ej <= Ei and Ei - Si > Ej - Sj, we say that cowi is stronger than cowj.
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 cowi.
Sample Input
3
1 2
0 3
3 4
0
Sample Output
1 0 0
Hint
Huge input and output,scanf and printf is recommended.
Source
POJ Contest,Author:Mathematica@ZSU

事实上，我感觉这道题出出来是考察我们的英文水平的。让我们用$OI$的语法来翻译一下题目，有$n$个区间，求出每个区间被包含的次数（重合不算）。显然先按照左端点排序，然后所有区间的左端点就是单调不减的了。这样的话答案就只与右端点有关了。这个直接区间查询单点插入就行了。

代码如下：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#define lc (p<<1)
#define rc (p<<1|1)
#define mid (T[p].l+T[p].r>>1)
#define N 100005
using namespace std;
inline int read(){
    int ans=0;
    char ch=getchar();
    while(!isdigit(ch))ch=getchar();
    while(isdigit(ch))ans=(ans<<3)+(ans<<1)+ch-'0',ch=getchar();
    return ans;
}
int n,ans[N];
struct Node{int s,t,id;}q[N];
struct st{int l,r,sum;}T[N<<2];
inline void pushup(int p){T[p].sum=T[lc].sum+T[rc].sum;}
inline void build(int p,int l,int r){
    T[p].l=l,T[p].r=r,T[p].sum=0;
    if(l==r)return;
    build(lc,l,mid);
    build(rc,mid+1,r);
}
inline void update(int p,int k){
    if(T[p].l==T[p].r){
        ++T[p].sum;
        return;
    }
    if(k<=mid)update(lc,k);
    else update(rc,k);
    pushup(p);
}
inline int query(int p,int ql,int qr){
    if(ql>T[p].r||T[p].l>qr)return 0;
    if(ql<=T[p].l&&T[p].r<=qr)return T[p].sum;
    if(qr<=mid)return query(lc,ql,qr);
    if(ql>mid)return query(rc,ql,qr);
    return query(lc,ql,mid)+query(rc,mid+1,qr);
}
inline bool cmp(Node a,Node b){return a.s==b.s?a.t>b.t:a.s<b.s;}
int main(){
    while(scanf("%d",&n)&&n){
        for(int i=1;i<=n;++i)q[i].s=read()+1,q[i].t=read()+1,q[i].id=i;
        build(1,1,100000);
        sort(q+1,q+n+1,cmp);
        for(int i=1;i<=n;++i){
            if(q[i].s==q[i-1].s&&q[i].t==q[i-1].t){
                ans[q[i].id]=ans[q[i-1].id];
            }
            else ans[q[i].id]=query(1,q[i].t,100000);
            update(1,q[i].t);
        }
        for(int i=1;i<=n;++i)printf("%d ",ans[i]);
        puts("");
    }
    return 0;
}