poj2481(线段树单点更新)

Cows

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

3
1 2
0 3
3 4
0

Sample Output

1 0 0

题意:给出n个区间的范围，求每一个区间是多少区间的真子集

思路:先对S从升序排序，如果S相等则对G进行降序排序，然后循环一遍查询再插入。这样只要统计当前区间的G到n范围里有多少其他区间的G，即得到答案。因为经过排序后node[i].s>=node[i-x].s;

#include <iostream>#include <fstream>#include <cstdio>#include <cstring>#include <queue>#include <stack>#include <vector>#include <map>#include <cmath>#include <algorithm>#include <functional>#define inf 0X3f3f3f3fusing namespace std;typedef long long ll;const int MAXN=1e9+10;const int MAX=100000+10;int n;struct COW{    int s,g,id;}cow[MAX];struct NODE{    int l,r,sum;}tree[MAX*4];int cmp(COW a,COW b){    if(a.s!=b.s)        return a.s<b.s;    else        return a.g>b.g;}int ans[MAX];void init(int node,int l,int r){    tree[node].l=l;    tree[node].r=r;    tree[node].sum=0;    if(l==r)        return ;    int mid=(l+r)>>1;    init(node<<1,l,mid);    init(node<<1|1,mid+1,r);}void updata_node(int node,int flag){    tree[node].sum++;    int l=tree[node].l;    int r=tree[node].r;    if(l==r)    return;    int mid=(l+r)>>1;    if(flag<=mid)   updata_node(node<<1,flag);    else    updata_node(node<<1|1,flag);}int query(int node,int x,int y){    int temp=0;    int l=tree[node].l;    int r=tree[node].r;    if(x<=l&&r<=y)        return tree[node].sum;    int mid=(l+r)>>1;    if(x<=mid)  temp+=query(node<<1,x,y);    if(y>mid)   temp+=query(node<<1|1,x,y);    return temp;}int main(){    #ifdef ONLINE_JUDGE    #else    freopen("in.txt","r",stdin);    #endif    while(cin>>n){        memset(ans,0,sizeof(ans));        if(n==0)            break;        for(int i=1;i<=n;i++){            scanf("%d%d",&cow[i].s,&cow[i].g);            cow[i].s++;cow[i].g++;            cow[i].id=i;        }        sort(cow+1,cow+n+1,cmp);        init(1,1,n);        for(int i=1;i<=n;i++){            if(i!=1&&cow[i].s==cow[i-1].s&&cow[i].g==cow[i-1].g)                ans[cow[i].id]=ans[cow[i-1].id];            else{                ans[cow[i].id]=query(1,cow[i].g,n);            }            cout<<query(1,cow[i].g,n)<<endl;            updata_node(1,cow[i].g);        }        for(int i=1;i<=n;i++){            if(i==1)                cout<<ans[i];            else            cout<<" "<<ans[i];        }        cout<<endl;    }    return 0;}