[树状数组] 树状数组基础练习题题解（二）

Time Limit: 3000MS
Memory Limit: 65536K

Contents

1 Description
2 Input
3 Output
4 Sample Input
5 Sample Output
6 Hint
7 Source

Description

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: cowi and cowj, their favourite clover range is [Si, Ei] and [Sj, Ej]. If Si <= Sj and Ej <= Ei and Ei – Si > Ej – Sj, we say that cowi is stronger than cowj.

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 <= 105), which is the number of cows. Then come N lines, the i-th of which contains two integers: S and E(0 <= S < E <= 105) 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 cowi.

Sample Input

Sample Output

1 0 0

Hint

Huge input and output,scanf and printf is recommended.

Source

POJ Contest,Author:Mathematica@ZSU

题目大意：有多少个区间包含区间 $$[l, r]$$

解题思路：将每一个区间按 $$r$$ 从大到小排序（这样可以保证区间编号小的 $$r$$ 一定大于等于区间编号大的 $$r$$ ），如果 $$r$$ 相等则按照l从小到大排序。一个区间 $$[l’, r’]$$ 包含另一个区间 $$[l, r]$$ 同时满足 $$l’ \leq l , r \leq r’$$ (不可同时取等号），排序后编号大的区间 $$r$$ 已经小于等于编号大的区间，现在我们可以通过树状数组快速求出小于等于 $$l$$ 的区间个数，输出即可，然后进行更新。

本题区间从 $$0$$ 开始，由于Lowbit的原因，我们将编号向右平移 $$1$$ 个单位

#include
#include
#include
#include
#include
#include
#define LL long long

using namespace std;

inline int read(){
    int x = 0;
    int p = 1;
    char ch = getchar();

    while(ch < '0' || ch > '9'){
        if(ch == '-')
            p = -1;
        ch = getchar();
    }

    while('0' <= ch && ch <= '9'){
        x = x*10 + ch - '0';
        ch = getchar();
    }

    return x*p;
}

const int MAXN = 100000;

struct Cow{
    int start;
    int end;
    int id;
};

Cow cow[MAXN+10];

int tree[MAXN + 10];
int ans[MAXN + 10];
int n;

inline bool cmp(Cow x,Cow y){
    if(x.end == y.end)
        return x.start < y.start;
    return x.end > y.end;
}

inline int lowbit(int x){
    return x & (-x);
}

inline void update(int i,int x){
    while(i<=MAXN){
        tree[i] += x;
        i += lowbit(i);
    }
}

inline int sum(int i){
    int ans = 0;

    while(i){
        ans += tree[i];
        i -= lowbit(i);
    }

    return ans;
}

int main(){

    while(scanf("%d",&n) && n){
        memset(cow,0,sizeof(cow));
        memset(ans,0,sizeof(ans));
        memset(tree,0,sizeof(tree));


        for(int i=0;i