POJ 2481 Cows

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.

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题意大意：FJ有n头牛(编号为1~n)，每一头牛都有一个测验值（S，E），对于牛i和牛j来说，如果它们的测验值满足下面的条件则表示牛i比牛j强壮：Si <= Sj and Ej <= Ei and Ei - Si > Ej - Sj。现在已知每一头牛的测验值，要求输出每头牛有几头牛比其强壮。
思路：首先将S[i]降序排序，如果相等则按E[i]的升序，根据排序后的顺序往每头牛的S位置加1。如果遇到完全重合的数据，则不需要重新进行求和计算，只需把前面的值赋值给它，否则需要计算获取到牛的S位置的元素的和。

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#include<algorithm>#include<iostream>#include<cstring>#include<cstdio>using namespace std;const int maxn=100005;int n,tmp,tr[maxn],ans[maxn];struct cow{    int s,e,y;}a[maxn],pre;bool cmp(cow c,cow d){    if(c.e==d.e)        return c.s<d.s;    return c.e>d.e;}int lowbit(int x){    return x&(-x);}int getsum(int pos){    int sum=0;    while(pos>0)    {        sum+=tr[pos];        pos-=lowbit(pos);    }    return sum;}void add(int pos){    while(pos<=100001)    {        tr[pos]++;        pos+=lowbit(pos);    }}int main(){    while(scanf("%d",&n)&&n)    {        memset(tr,0,sizeof(tr));        pre.s=pre.e=0;        for(int i=1;i<=n;i++)        {            scanf("%d%d",&a[i].s,&a[i].e);            a[i].s++,a[i].e++,a[i].y=i;        }        sort(a+1,a+n+1,cmp);        for(int i=1;i<=n;i++)        {            if(a[i].s!=pre.s||a[i].e!=pre.e)            {                tmp=ans[a[i].y]=getsum(a[i].s);                pre=a[i];            }            else                ans[a[i].y]=tmp;            add(a[i].s);        }        for(int i=1;i<=n;i++)            printf("%d ",ans[i]);        printf("\n");    }    return 0;}