POJ 2352 Stars （线段树&&树状数组）

Stars

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 32475 Accepted: 14186

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

The first line of the input file contains a number of stars N (1<=N<=15000). The following N lines describe coordinates of stars (two integers X and Y per line separated by a space, 0<=X,Y<=32000). There can be only one star at one point of the plane. Stars are listed in ascending order of Y coordinate. Stars with equal Y coordinates are listed in ascending order of X coordinate. 

Output

The output should contain N lines, one number per line. The first line contains amount of stars of the level 0, the second does amount of stars of the level 1 and so on, the last line contains amount of stars of the level N-1.

Sample Input

51 15 17 13 35 5

Sample Output

12110

Hint

This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.

学习树状数组的第一题，很棒上周周赛的A题，刚学线段树，知道是个线段树的题目，但还是不会，研究一个周的线段树了，希望能有点起色

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <cmath>#define init(a) memset(a,0,sizeof(a))using namespace std;#define MAX INT_MAX#define MIN INT_MIN#define LL __int64#define lson l , m , rt << 1#define rson m + 1 , r , rt << 1 | 1const int maxn = 50010;using namespace std;int c[maxn];int lev[maxn];int lowbit(int x){    return x&(-x);}void add(int i,int w){    while(i<=32010)    {        c[i]+=w;        i+=lowbit(i);    }}int sum(int i){    int s=0;    while(i>0)    {        s += c[i];        i =i - lowbit(i);    }    return s;}int main(){    int x,y,n;    while(scanf("%d",&n)!=EOF)    {        init(c);        init(lev);        for(int i=0;i<n;i++)        {            scanf("%d%d",&x,&y);            int temp=sum(x+1);            lev[temp]++;            add(x+1,1);        }        for(int i=0;i<n;i++)         printf("%d\n",lev[i]);    }    return 0;}

线段树代码

单点更新 区间查询 简单哈希

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <cmath>#define init(a) memset(a,0,sizeof(a))using namespace std;#define MAX INT_MAX#define MIN INT_MIN#define LL __int64#define lson l , m , rt << 1#define rson m + 1 , r , rt << 1 | 1const int maxn = 50010;struct node{int left,right;int num; //}T[4*maxn];int hash[maxn];int L[maxn];int ans=0,M;void Creat(int left,int right,int id)//建树{T[id].left =left;T[id].right =right;T[id].num =0;if(T[id].left ==T[id].right )return ;Creat(left,(left+right)/2,2*id);Creat((left+right)/2+1,right,2*id+1);}void UPdata(int id,int i,int j){if(T[id].left<=i&&T[id].right >=i)T[id].num ++;if(T[id].left ==T[id].right )return;if(i>T[id].right )return;if(i<T[id].left )return;int mid=(T[id].left +T[id].right )/2;if(i<=mid)UPdata(id*2,i,j);elseUPdata(id*2+1,i,j);}void query(int id,int l,int r){int mid=(T[id].left +T[id].right)/2;if(T[id].left ==l&&T[id].right ==r){ans+=T[id].num ;return;}if(r<=mid)query(2*id,l,r);else if(l>mid)query(2*id+1,l,r);else{query(2*id,l,mid);query(2*id+1,mid+1,r);}}int main(){int n,r;while(scanf("%d",&n)!=EOF)        {        M = 0;for(int i=0;i<n;i++){scanf("%d%d",&L[i],&r);if(L[i]>M)                M = L[i];}    Creat(0,M,1);            for(int i = 0;i<n;i++)            {                ans = 0;                query(1,0,L[i]);                hash[ans]++;   UPdata(1,L[i],1);            }            for(int i = 0;i<n;i++)                printf("%d\n",hash[i]);        }return 0;}