Cows
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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!
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!
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.
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.
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.
31 20 33 40
1 0 0
Hint
Huge input and output,scanf and printf is recommended.
题解:将区间按r的降序,l的升序排序,然后仿照poj2352 stars。
#include<iostream>#include<stdio.h>#include<algorithm>#include<string.h>using namespace std;const int Maxn=100005;int sum[Maxn],ans[Maxn];int n;struct node{ int l,r,index;}a[Maxn];bool cmp(node a,node b){ if(a.r==b.r) return a.l<b.l; return a.r>b.r;}int lowbit(int x){ return x&(-x);}void updata(int x){ while(x<=Maxn){ sum[x]++; x+=lowbit(x); }}int getSum(int x){ int s=0; while(x>0){ s+=sum[x]; x-=lowbit(x); } return s;}int main(){ //freopen("in.txt","r",stdin); while(scanf("%d",&n)&&n){ memset(sum,0,sizeof(sum)); for(int i=1;i<=n;i++) scanf("%d%d",&a[i].l,&a[i].r),a[i].l++,a[i].r++,a[i].index=i; sort(a+1,a+1+n,cmp); ans[a[1].index]=0; for(int i=2;i<=n;i++) if(a[i].l==a[i-1].l&&a[i].r==a[i-1].r) ans[a[i].index]=ans[a[i-1].index]-1; else ans[a[i].index]=0; for(int i=1;i<=n;i++){ ans[a[i].index]+=getSum(a[i].l); updata(a[i].l); } for(int i=1;i<n;i++) printf("%d ",ans[i]); printf("%d\n",ans[n]); }}
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