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Time Limit: 20 Sec
Memory Limit: 256 MB
codeforces.com/problemset/problem/91/B
Input
The first line contains an integer n (2 ≤ n ≤ 105) — the number of walruses in the queue. The second line contains integers ai (1 ≤ ai ≤ 109).
Note that some walruses can have the same age but for the displeasure to emerge the walrus that is closer to the head of the queue needs to be strictly younger than the other one.
Output
Print n numbers: if the i-th walrus is pleased with everything, print "-1" (without the quotes). Otherwise, print the i-th walrus‘s displeasure: the number of other walruses that stand between him and the furthest from him younger walrus.
Sample Input
6
10 8 5 3 50 45
Sample Output
2 1 0 -1 0 -1
题意
给你一个数列，让你找到最右边比这个数小的数的位置，如果没有就输出-1
题解：
直接线段树中二分，查询最小值，然后二分区间就好了~
代码
//qscqesze #include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <vector> #include <sstream> #include <queue> #include <typeinfo> #include <fstream> #include <map> #include <stack> typedef long long ll; using namespace std; //freopen("D.in","r",stdin); //freopen("D.out","w",stdout); #define sspeed ios_base::sync_with_stdio(0);cin.tie(0) #define maxn 2000001 #define mod 1000000007 #define eps 1e-9 int Num; char CH[20]; const int inf=0x3f3f3f3f; inline ll read() { int x=0,f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} return x*f; } //************************************************************************************** int tr[maxn]; int a[maxn]; int ans[maxn]; int tmp; void build(int x,int l,int r) { if(l==r) { tr[x]=a[l]; return; } int mid=(l+r)>>1; build(x<<1,l,mid); build(x<<1|1,mid+1,r); tr[x]=min(tr[x<<1],tr[x<<1|1]); } void query(int x,int l,int r,int t) { if(l==r) { ans[tmp++]=l-t-1; return; } int mid=(l+r)>>1; if(tr[x<<1|1]<a[t]) query(x<<1|1,mid+1,r,t); else query(x<<1,l,mid,t); } void update(int x,int l,int r,int t) { if(l==r) { tr[x]=inf; return; } int mid=(l+r)>>1; if(t<=mid) update(x<<1,l,mid,t); else update(x<<1|1,mid+1,r,t); tr[x]=min(tr[x<<1],tr[x<<1|1]); } int main() { int n=read(); for(int i=1;i<=n;i++) a[i]=read(); build(1,1,n); for(int i=1;i<=n;i++) { if(tr[1]>=a[i]) ans[tmp++]=-1; else query(1,1,n,i); update(1,1,n,i); } for(int i=0;i<n;i++) printf("%d ",ans[i]); }
Codeforces Beta Round #75 (Div. 1 Only) B. Queue 线段树+二分
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原文地址：http://www.cnblogs.com/qscqesze/p/4604355.html
