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题意:
n个牛,每个牛对应一个区间,对于每个牛求n个区间有几个包含该牛的区间。
分析:
先 区间右边界从大到小排序,相同时左边界小到大,统计第i头牛即左边界在前i-1头左边界的正序数。
#include <map> #include <set> #include <list> #include <cmath> #include <queue> #include <stack> #include <cstdio> #include <vector> #include <string> #include <cctype> #include <complex> #include <cassert> #include <utility> #include <cstring> #include <cstdlib> #include <iostream> #include <algorithm> using namespace std; typedef pair<int,int> PII; typedef long long ll; #define lson l,m,rt<<1 #define pi acos(-1.0) #define rson m+1,r,rt<<11 #define All 1,N,1 #define read freopen("in.txt", "r", stdin) #define N 100010 const ll INFll = 0x3f3f3f3f3f3f3f3fLL; const int INF= 0x7ffffff; const int mod = 1000000007; int bit[N],tmp[N],n; struct node{ int s,e,index; }a[N]; bool cmp(node x,node y){ if(y.e==x.e) return x.s<y.s; else return x.e>y.e; } int sum(int x){ int num=0; while(x>0){ num+=bit[x]; x-=(x&(-x)); } return num; } void add(int x,int d){ while(x<=N){ bit[x]+=d; x+=(x&(-x)); } } void solve(){ memset(bit,0,sizeof(bit)); sort(a,a+n,cmp); tmp[a[0].index]=0; add(a[0].s,1); for(int i=1;i<n;++i){ if(a[i].s==a[i-1].s&&a[i].e==a[i-1].e){ tmp[a[i].index]=tmp[a[i-1].index]; } else{ // cout<<i<<" "<<sum(a[i].e)<<endl; tmp[a[i].index]=sum(a[i].s); } add(a[i].s,1); } for(int i=0;i<n;++i){ if(i==n-1) printf("%d\n",tmp[i]); else printf("%d ",tmp[i]); } } int main() { while(~scanf("%d",&n)){ if(n==0)break; for(int i=0;i<n;++i){ scanf("%d%d",&a[i].s,&a[i].e); a[i].s++; a[i].e++; a[i].index=i; } solve(); } return 0; }
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原文地址:http://www.cnblogs.com/zsf123/p/4716466.html