| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 10785 | Accepted: 3572 |
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!
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.
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
3 1 2 0 3 3 4 0
Sample Output
1 0 0
1 #include <stdio.h> 2 #include <string.h> 3 #include <math.h> 4 #include <algorithm> 5 #include <iomanip> 6 #include <set> 7 #include <map> 8 #include <vector> 9 #include <queue> 10 #define llt long long int 11 #define ml(x, y) ((x + y) >> 1) 12 #define mr(x, y) (((x + y) >> 1) + 1) 13 #define pl(p) (p << 1) 14 #define pr(p) ((p << 1) | 1) 15 #define N 100005 16 using namespace std; 17 struct node 18 { 19 int s, e, pos; 20 bool operator < (const node& a)const//按照先左边小，再右边大进行排序 21 { 22 return s == a.s ? e > a.e : s < a.s; 23 } 24 }a[N]; 25 int segtree[N << 2], ans[N]; 26 void build(int l, int r, int p) 27 { 28 segtree[p] = 0; 29 if (l < r) 30 { 31 build(l, ml(l, r), pl(p)); 32 build(mr(l, r), r, pr(p)); 33 } 34 } 35 void update(int l, int r, int p, int v) 36 { 37 segtree[p]++; 38 if (l != r) 39 if (ml(l, r) >= v) 40 update(l, ml(l, r), pl(p), v); 41 else 42 update(mr(l, r), r, pr(p), v); 43 } 44 int query(int l, int r, int p, int v)//查询区间[l, imax]的个数 45 { 46 if (v <= l) 47 return segtree[p]; 48 int ans = 0; 49 if (ml(l, r) >= v)//等于[l,mid]+[mid+1,imax] 50 ans = segtree[pr(p)] + query(l, ml(l, r), pl(p), v); 51 else//[mid+1,imax] 52 ans = query(mr(l, r), r, pr(p), v); 53 return ans; 54 } 55 int main() 56 { 57 int n, i, maxe;//maxe记录一个最大值 58 while (scanf("%d", &n), n) 59 { 60 maxe = 0; 61 for (i = 0; i < n; i++) 62 { 63 scanf("%d%d", &a[i].s, &a[i].e); 64 a[i].pos = i; 65 if (maxe < a[i].e) 66 maxe = a[i].e; 67 } 68 sort(a, a + n);//排序 69 build(0, maxe, 1); 70 for (i = 0; i < n; i++) 71 { 72 if (i != 0 && a[i].e == a[i - 1].e && a[i].s == a[i - 1].s)//处理重复的 73 ans[a[i].pos] = ans[a[i - 1].pos];// 74 else 75 ans[a[i].pos] = query(0, maxe, 1, a[i].e);//查询只需要得要a[i].e到imax有多少个数就行了 76 update(0, maxe, 1, a[i].e);//因为s小的先处理,所以只需要根据e来更新，记录该区间e的个数 77 } 78 printf("%d", ans[0]); 79 for (i = 1; i < n; i++) 80 printf(" %d", ans[i]); 81 putchar(' '); 82 } 83 return 0; 84 }