Why Did the Cow Cross the Road III

時間限制:?1 Sec??內存限制:?128 MB
提交:?65??解決:?28
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題目描述

The layout of Farmer John's farm is quite peculiar, with a large circular road running around the perimeter of the main field on which his cows graze during the day. Every morning, the cows cross this road on their way towards the field, and every evening they all cross again as they leave the field and return to the barn.?
As we know, cows are creatures of habit, and they each cross the road the same way every day. Each cow crosses into the field at a different point from where she crosses out of the field, and all of these crossing points are distinct from each-other. Farmer John owns?N?cows, conveniently identified with the integer IDs?1…N, so there are precisely?2N?crossing points around the road. Farmer John records these crossing points concisely by scanning around the circle clockwise, writing down the ID of the cow for each crossing point, ultimately forming a sequence with?2N?numbers in which each number appears exactly twice. He does not record which crossing points are entry points and which are exit points.

Looking at his map of crossing points, Farmer John is curious how many times various pairs of cows might cross paths during the day. He calls a pair of cows?(a,b)?a "crossing" pair if cow?a's path from entry to exit must cross cow?b's path from entry to exit. Please help Farmer John count the total number of crossing pairs.

輸入

The first line of input contains?N?(1≤N≤50,000), and the next?2N?lines describe the cow IDs for the sequence of entry and exit points around the field.

輸出

Please print the total number of crossing pairs.

樣例輸入

4 3 2 4 4 1 3 2 1

樣例輸出

3
【題意】在一個圓上，順時針 給出一些數字，每個數字出現兩遍，然后數字相同的連邊，問多少對邊相交。
【分析】我們可以發現按照題目給出的數據順序，如果兩個數的連線相交，那么他倆在一維中的連線一相交，也就是如果某個數兩次出現
的位置中間有多少個數 出現了一次，那么就有多少條線與他相交，那我們直接對于每個數統計兩個位置之間的出現一次的數就行了。
對于這種數據量較大無法N方解決的統計問題，樹狀數組一般都可以。對于這個題，從左到右掃，當第一次掃到這個數時，從這個位置
向上lowbit依次+1，當第二次掃到這個數的時候，統計第一次出現的位置到當前位置的sum值，然后從第一次出現的位置向上lowbit
依次-1，表示刪除此邊。
#include <bits/stdc++.h> #define inf 0x3f3f3f3f #define met(a,b) memset(a,b,sizeof a) #define pb push_back #define mp make_pair #define inf 0x3f3f3f3f using namespace std; typedef long long ll; const int N = 1e5+50; const int mod = 1e9+7; const double pi= acos(-1.0); typedef pair<int,int>pii; int n,ans; int a[N],sum[N],vis[N]; void upd(int x,int add){for(int i=x;i<=2*n;i+=i&(-i)){sum[i]+=add;} } int qry(int x){int ret=0;for(int i=x;i>=1;i-=i&(-i)){ret+=sum[i];}return ret; } int main(){scanf("%d",&n);for(int i=1;i<=2*n;i++){scanf("%d",&a[i]);}for(int i=1;i<=2*n;i++){if(!vis[a[i]]){upd(i,1);vis[a[i]]=i;}else {int s=qry(i-1)-qry(vis[a[i]]);//printf("i:%d ai:%d l:%d r:%d\n",i,a[i],qry(vis[a[i]]),qry(i-1));ans+=s;upd(vis[a[i]],-1);}}printf("%d\n",ans);return 0; }

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轉載于:https://www.cnblogs.com/jianrenfang/p/7253872.html

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