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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.
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.
Print n integers in one line. For each x from 1 to n, print the maximum strength among all groups of size x.
10 1 2 3 4 5 4 3 2 1 6
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）的时候能够切割成一些区间。这每一个区间都会有一个最小值。在相同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，跟如今入栈的数之间相隔了多少个数，由于要求的是连续区间。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[N],ans[N]; 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实用了，能够出栈 int ls=k.width+len;//出栈的同一时候获得其长度 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--)//由于上面仅仅更新了一部分的点，所以如今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; }
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Codeforces548D:Mike and Feet(单调栈)
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原文地址：http://www.cnblogs.com/mengfanrong/p/4906494.html
