Codeforces548D:Mike and Feet(单调栈)

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.InputThe 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.OutputPrint 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

题意：

给出n个数，这n个数在区间长度为i（1~n）的时候可以分割成一些区间，这每个区间都会有一个最小值，在同样长度的这些区间的最小值中，输出最大值

思路：

使用单调栈，保持栈内的数的单调递增性

#include <iostream>#include <stdio.h>#include <string.h>#include <stack>#include <queue>#include <map>#include <set>#include <vector>#include <math.h>#include <bitset>#include <algorithm>#include <climits>using namespace std;#define LS 2*i#define RS 2*i+1#define UP(i,x,y) for(i=x;i<=y;i++)#define DOWN(i,x,y) for(i=x;i>=y;i--)#define MEM(a,x) memset(a,x,sizeof(a))#define W(a) while(a)#define gcd(a,b) __gcd(a,b)#define LL long long#define N 200005#define MOD 1000000007#define INF 0x3f3f3f3f#define EXP 1e-8#define lowbit(x) (x&-x)/*num栈顶的元素，width宽度，跟现在入栈的数之间相隔了多少个数，因为要求的是连续区间，所以这个也是必须记录的*/struct node{    int num,width;    node() {};    node(int _num,int _width):num(_num),width(_width) {}};stack<node> S;int a,ans;int main(){    int n,i,j;    scanf("%d",&n);    for(i = 0; i<n; i++)        scanf("%d",&a[i]);    a[n++] = 0;    MEM(ans,0);    for(i = 0; i<n; i++)    {        int len = 0;//连续区间的长度        node k;        while(!S.empty())        {            k = S.top();            if(k.num<a[i])                break;                //新入栈的元素比栈顶元素要小，那么对于这个连续区间而言，这个比新入栈的元素就没有用了，可以出栈            int ls=k.width+len;//出栈的同时获得其长度            if(k.num>ans[ls])//ans记录ls区间的时候的最大值            {                ans[ls]=k.num;            }            len+=k.width;            S.pop();        }        S.push(node(a[i],len+1));    }    for(i = n-1; i>=1; i--)//因为上面只更新了一部分的点，所以现在要对那些没有更新的点也更新        ans[i]=max(ans[i],ans[i+1]);    printf("%d",ans[1]);    for(i = 2; i<n; i++)        printf(" %d",ans[i]);    printf("\n");    return 0;}