Lost Cows
Description
N (2 <= N <= 8,000) cows have unique
brands in the range 1..N. In a spectacular display of poor judgment,
they visited the neighborhood 'watering hole' and drank a few too many
beers before dinner. When it was time to line up for their evening
meal, they did not line up in the required ascending numerical order of
their brands.
Regrettably, FJ does not have a way to sort them. Furthermore, he's
not very good at observing problems. Instead of writing down each
cow's brand, he determined a rather silly statistic: For each cow in
line, he knows the number of cows that precede that cow in line that do,
in fact, have smaller brands than that cow.
Given this data, tell FJ the exact ordering of the cows.
Input
* Line
1: A single integer, N
* Lines 2..N: These N-1 lines describe the number of cows that
precede a given cow in line and have brands smaller than that cow. Of
course, no cows precede the first cow in line, so she is not listed.
Line 2 of the input describes the number of preceding cows whose brands
are smaller than the cow in slot #2; line 3 describes the number of
preceding cows whose brands are smaller than the cow in slot #3; and so
on.
Output
* Lines
1..N: Each of the N lines of output tells the brand of a cow in line.
Line #1 of the output tells the brand of the first cow in line; line 2
tells the brand of the second cow; and so on.
Sample
Input
5
1
2
1
0
Sample Output
2
4
5
#include<stdio.h>
#define maxn 8000
int tree[maxn*3], s[maxn], ans[maxn];
void build(int l, int r, int depth)
{
tree[depth] = r - l + 1;
if (l != r)
{
int mid = (l + r) >> 1;
build(l, mid, (depth << 1));
build(mid + 1, r, (depth << 1) + 1);
}
}
void find(int l, int r, int index, int num, int depth)
{
tree[depth]--;
if (l == r)
{
ans[index] = r;
return;
}
else
{
int mid = (l + r) >> 1, m = depth << 1;
if (num <= tree[m])
{
find(l, mid, index, num, m);
}
else
{
num -= tree[m];
find(mid + 1, r, index, num, m + 1);
}
}
}
int main()
{
int n, i;
while (scanf("%d", &n) != EOF)
{
build(1, n, 1);
for (i = 2, s[1] = 1; i <= n; i++)
{
scanf("%d", &s[i]);
s[i]++;
}
for (i = n; i >= 1; i--)
{
find(1, n, i, s[i], 1);
}
for (i = 1; i <= n; i++)
{
printf("%d\n", ans[i]);
}
}
return 0;
}
3
1
題意：給出每個數(shù)所在位置的逆序列個數(shù)，求這些數(shù)。
用線段樹寫，一來統(tǒng)計個數(shù)，而來可以隨時刪除已去掉的數(shù)，這些操作都是logn。