poj 2352 Stars

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Description

Astronomers often examine star maps where stars are represented by points on a plane and each star has Cartesian coordinates. Let the level of a star be an amount of the stars that are not higher and not to the right of the given star. Astronomers want to know the distribution of the levels of the stars. 

For example, look at the map shown on the figure above. Level of the star number 5 is equal to 3 (it‘s formed by three stars with a numbers 1, 2 and 4). And the levels of the stars numbered by 2 and 4 are 1. At this map there are only one star of the level 0, two stars of the level 1, one star of the level 2, and one star of the level 3. 

You are to write a program that will count the amounts of the stars of each level on a given map.

Input

Output

Sample Input

5
1 1
5 1
7 1
3 3
5 5

Sample Output

1
2
1
1
0

Hint

题解及代码：

        水题，直接上代码。

树状数组：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#define maxn 32010
using namespace std;
typedef long long ll;

int level[maxn];
int c[maxn];

int lowbit(int i)
{
    return i&(-i);
}

int sum(int i)
{
    int ans=0;
    while(i>0)
    {
        ans+=c[i];
        i-=lowbit(i);
    }
    return ans;
}

void update(int i)
{
    while(i<=32001)
    {
        c[i]++;
        i+=lowbit(i);
    }
}


int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        int x,y;
        memset(c,0,sizeof(c));
        memset(level,0,sizeof(level));
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&x,&y);
            level[sum(x+1)]++;
            update(x+1);
        }
        for(int i=0;i<n;i++)
        {
            printf("%d\n",level[i]);
        }
    }
    return 0;
}

线段树：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#define maxn 32010
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
typedef long long ll;

int level[maxn];

struct segment
{
   int l,r;
   int value;
}son[maxn<<2];

void PushUp(int rt)
{
    son[rt].value=son[rt<<1].value+son[rt<<1|1].value;
}

void Build(int l,int r,int rt)
{
    son[rt].l=l;
    son[rt].r=r;
    son[rt].value=0;
    if(l==r) return;
    int m=(l+r)/2;
    Build(lson);
    Build(rson);
}

void Update(int p,int rt)
{
    if(son[rt].l==son[rt].r)
    {
        son[rt].value++;
        return;
    }

    int m=(son[rt].l+son[rt].r)/2;
    if(p<=m)
        Update(p,rt<<1);
    else Update(p,rt<<1|1);

    PushUp(rt);
}

int Query(int l,int r,int rt)
{
    if(son[rt].l==l&&son[rt].r==r)
    {
        return son[rt].value;
    }

    int m=(son[rt].l+son[rt].r)/2;
    int ret=0;

    if(r<=m)
        ret=Query(l,r,rt<<1);
    else if(l>m)
        ret=Query(l,r,rt<<1|1);
    else{
        ret=Query(lson);
        ret+=Query(rson);
    }
    return ret;
}

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        Build(1,32001,1);
        int x,y;
        //memset(c,0,sizeof(c));
        memset(level,0,sizeof(level));
        for(int i=0;i<n;i++)
        {
            scanf("%d%d",&x,&y);
            //cout<<Query(1,x+1,1)<<endl;;
            level[Query(1,x+1,1)]++;
            Update(x+1,1);
        }
        for(int i=0;i<n;i++)
        {
            printf("%d\n",level[i]);
        }
    }
    return 0;
}

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poj 2352 Stars

标签：des   style   http   color   os   io   strong   for   

原文地址：http://blog.csdn.net/knight_kaka/article/details/38612675