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Time Limit: 20 Sec
Memory Limit: 256 MB
http://codeforces.com/problemset/problem/319/B
Input
The first line of input contains integer n denoting the number of psychos, (1 ≤ n ≤ 105). In the second line there will be a list of n space separated distinct integers each in range 1 to n, inclusive — ids of the psychos in the line from left to right.
Output
Print the number of steps, so that the line remains the same afterward..
Sample Input
10
10 9 7 8 6 5 3 4 2 1
Sample Output
2
题意
给你一堆数，每个数次都可以吃掉他右边连续减小的序列，问你多少轮之后，会变成上升序列
题解：
一开始想的暴力，用并查集/链表优化一下，结果不幸被t（其实也是情理之中……
T的数据是 100000 1 2 3 4 5…… 99999
正解是单调队列，维护一个单调下降的序列，就可以找到一直杀到哪儿的位置。
f[i]表示杀人的次数
代码
#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 test freopen("test.txt","r",stdin) #define maxn 1000005 #define mod 10007 #define eps 1e-5 const int inf=0x3f3f3f3f; const ll infll = 0x3f3f3f3f3f3f3f3fLL; inline ll read() { ll 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 a[maxn]; int f[maxn]; int t[maxn]; int main() { int n=read(),top=0,ans=0; for(int i=0;i<n;i++) a[i]=read(); for(int i=n-1;i>=0;i--) { int tt=0; while(top&&a[t[top-1]]<a[i]) f[i]=tt=max(tt+1,f[t[--top]]); t[top++]=i; } printf("%d",*max_element(f,f+n)); }
Codeforces Round #189 (Div. 1) B. Psychos in a Line 单调队列
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原文地址：http://www.cnblogs.com/qscqesze/p/4609626.html
