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  • CodeForces

    Nested Segments

    You are given n segments on a line. There are no ends of some segments that coincide. For each segment find the number of segments it contains.

    Input

    The first line contains a single integer n (1 ≤ n ≤ 2·105) — the number of segments on a line.

    Each of the next n lines contains two integers li and ri ( - 109 ≤ li < ri ≤ 109) — the coordinates of the left and the right ends of the i-th segment. It is guaranteed that there are no ends of some segments that coincide.

    Output

    Print n lines. The j-th of them should contain the only integer aj — the number of segments contained in the j-th segment.

    Examples

    Input
    4
    1 8
    2 3
    4 7
    5 6
    Output
    3
    0
    1
    0
    Input
    3
    3 4
    1 5
    2 6
    Output
    0
    1
    1

    题意:给出n条线段,求线段包含情况,输出包含数。
    思路:使用树状数组,数据太大,需要离散化。然后按左端点排序即可。
    #include<bits/stdc++.h>
    using namespace std;
    const int maxn=400010;
    int a[maxn],tree[maxn],n,ans[maxn];
    struct node{
        int l,r,id;
        bool friend operator < (const node &u,const node &v){
            if(u.l==v.l) return u.r<v.r;
            return u.l>v.l;
        }
    } p[maxn];
    int lowbit(int k){
        return k&(-k);
    }
    void updata(int k){
        while(k<maxn){
            tree[k]+=1;
            k+=lowbit(k);
        }
    }
    int sum(int k){
        int ans=0;
        while(k){
            ans+=tree[k];
            k-=lowbit(k);
        }
        return ans;
    }
    int main(){
        int cnt=0;
        scanf("%d",&n);
        for(int i=1; i<=n; i++){
            scanf("%d %d",&p[i].l,&p[i].r);
            a[++cnt]=p[i].r;
            p[i].id=i;
        }
        sort(a+1,a+1+cnt);
        cnt=unique(a+1,a+cnt+1)-(a+1);
        for(int i=1; i<=n; i++) p[i].r=lower_bound(a+1,a+1+cnt,p[i].r)-a;
        sort(p+1,p+1+n);
        for(int i=1; i<=n; i++){
            ans[p[i].id]=sum(p[i].r+1);
            updata(p[i].r+1);
        }
        for(int i=1; i<=n; i++)
            printf("%d
    ",ans[i]);
    }
    View Code

    PS:摸鱼怪的博客分享,欢迎感谢各路大牛的指点~

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  • 原文地址:https://www.cnblogs.com/MengX/p/9079949.html
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