Codeforces Round #305 (Div. 1) B. Mike and Feet
2015-07-13 20:16
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Mike is the president of country What-The-Fatherland. There are n bears living in this country besides Mike. All of them are standing
in a line and they are numbered from 1 to n from
left to right. i-th bear is exactly ai feet
high.
A group of bears is a non-empty contiguous segment of the line. The size of a group is the number of bears in that group. The strengthof
a group is the minimum height of the bear in that group.
Mike is a curious to know for each x such that 1 ≤ x ≤ n the
maximum strength among all groups of size x.
Input
The first line of input contains integer n (1 ≤ n ≤ 2 × 105),
the number of bears.
The second line contains n integers separated by space, a1, a2, ..., an (1 ≤ ai ≤ 109),
heights of bears.
Output
Print n integers in one line. For each x from 1 to n,
print the maximum strength among all groups of size x.
Sample test(s)
input
output
6 4 4 3 3 2 2 1 1 1
这题可以用单调栈做,维护一个栈,记录minmum(该区间的最小值)和count(区间的总长度)。
in a line and they are numbered from 1 to n from
left to right. i-th bear is exactly ai feet
high.
A group of bears is a non-empty contiguous segment of the line. The size of a group is the number of bears in that group. The strengthof
a group is the minimum height of the bear in that group.
Mike is a curious to know for each x such that 1 ≤ x ≤ n the
maximum strength among all groups of size x.
Input
The first line of input contains integer n (1 ≤ n ≤ 2 × 105),
the number of bears.
The second line contains n integers separated by space, a1, a2, ..., an (1 ≤ ai ≤ 109),
heights of bears.
Output
Print n integers in one line. For each x from 1 to n,
print the maximum strength among all groups of size x.
Sample test(s)
input
10 1 2 3 4 5 4 3 2 1 6
output
6 4 4 3 3 2 2 1 1 1
这题可以用单调栈做,维护一个栈,记录minmum(该区间的最小值)和count(区间的总长度)。
#include<iostream> #include<stdio.h> #include<stdlib.h> #include<string.h> #include<algorithm> using namespace std; #define maxn 200060 int ans[maxn]; struct node{ int count,minmum; }stack[maxn]; int main() { int n,m,i,j,top,count,b; while(scanf("%d",&n)!=EOF) { memset(ans,0,sizeof(ans)); top=0; for(i=1;i<=n;i++){ scanf("%d",&b); count=0; while(top>0 && stack[top].minmum>=b){ stack[top].count+=count; count=stack[top].count; if(ans[count]<stack[top].minmum){ ans[count]=stack[top].minmum; } top--; } top++; stack[top].minmum=b; stack[top].count=count+1; } count=0; while(top>0){ stack[top].count+=count; count=stack[top].count; if(ans[count]<stack[top].minmum){ ans[count]=stack[top].minmum; } top--; } for(i=n;i>=2;i--){ if(ans[i]>ans[i-1]){/*这里算出来的ans[i]是连续长度为i的区间的最小值,但这个最小值是所有连续长度为i的区间长度的最大值,下面如果ans[i+1]比ans[i]大,那么ans[i]可以更新为ans[i+1],因为如果i+1个连续数区间的最小值的最大值是b,那么去掉一个数,一定可以做到长度为i的连续数区间的最大值是b。*/ ans[i-1]=ans[i]; } } for(i=1;i<=n-1;i++){ printf("%d ",ans[i]); } printf("%d\n",ans[i]); } return 0; }
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