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
3
1
Source

USACO 2003 U S Open Orange
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分析

樹狀數組的 SUM( X ) 用于記錄 編號X 后面滿足小于等于 X 的已經用掉了的編號的個數；F[ i ] 就是題目給出的 第 i 個牛 前面比 第i 個牛的編號小的編號的個數； 我們需要二分的就是 X，判斷 X是不是當前第 i 頭牛的編號。

如果 (X-1)-SUM( X - 1) == F[ i ]
(即 編號X 前面剩下的小于 X 的編號的數量恰好等于 第 i 頭牛編號的條件， 則 X 就是 第 i 頭牛的編號啦）

如果(X-1)-SUM( X - 1) > F[ i ]
（說明編號偏大咯）

如果(X-1)-SUM( X - 1) < F[ i ]
(說明編號偏小咯）
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程序：

#include <iostream> #include <algorithm> #include <cstdio> #include <cstring> #include <cmath> #include <queue> using namespace std; int n,tree[10000],f[10000],num[10000]; int lowbit(int x) {return x & -x; } void add(int x,int k) {while(x<=n){tree[x]+=k;x+=lowbit(x);} } int sum(int x) {int ans=0;while(x>0){ans+=tree[x];x-=lowbit(x);}return ans; } int main() {scanf("%d",&n);num[1]=0;for (int i=2;i<=n;i++)scanf("%d",&f[i]);num[n]=f[n]+1;add(num[n],1);for (int i=n-1;i>0;i--){int l=1,r=n;while (r>l){int mid=(l+r)>>1;if (mid-1-sum(mid)>=f[i]) r=mid; else l=mid+1;}num[i]=l;add(num[i],1);}for (int i=1;i<=n;i++)printf("%d\n",num[i]);return 0; }

轉載于:https://www.cnblogs.com/YYC-0304/p/10292858.html

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