POJ 2559 Largest Rectangle in a Histogram(单调栈）

Description

A histogram is a polygon composed of a sequence of rectangles aligned at a common base line. The rectangles have equal widths but may have different heights. For example, the figure on the left shows the histogram that consists of rectangles with the heights 2, 1, 4, 5, 1, 3, 3, measured in units where 1 is the width of the rectangles: 

Usually, histograms are used to represent discrete distributions, e.g., the frequencies of characters in texts. Note that the order of the rectangles, i.e., their heights, is important. Calculate the area of the largest rectangle in a histogram that is aligned at the common base line, too. The figure on the right shows the largest aligned rectangle for the depicted histogram.

Input

Output

Sample Input

7 2 1 4 5 1 3 3
4 1000 1000 1000 1000
0

Sample Output

8
4000

学习了下单调栈写法。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
typedef long long LL;
typedef pair<int,int>pil;
const int INF = 0x3f3f3f3f;
const int maxn=1e5+100;
struct node{
    int val;
    int pos;
};
stack<node>s;
int n,a[maxn];
int main()
{
    node st;
    while(~scanf("%d",&n)&&n)
    {
        REPF(i,1,n)
          scanf("%d",&a[i]);
        st.val=st.pos=0;
        s.push(st);LL ans=0;
        for(int i=1;i<=n;i++)
        {
            st.pos=i;
            while(s.top().val>a[i])
            {
                st=s.top();
                s.pop();
                LL temp=(LL)st.val*(i-st.pos);
                ans=max(ans,temp);
            }
            st.val=a[i];
            s.push(st);
        }
        while(!s.empty())
        {
            st=s.top();
            LL temp=(LL)st.val*(n-st.pos+1);
            ans=max(ans,temp);
            s.pop();
        }
        printf("%I64d\n",ans);
    }
    return 0;
}


/*

*/

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
typedef long long LL;
typedef pair<int,int>pil;
const int INF = 0x3f3f3f3f;
const int maxn=1e5+100;
int l[maxn],r[maxn];
int n,num[maxn];
int main()
{
    while(~scanf("%d",&n)&&n)
    {
        REPF(i,1,n)
            scanf("%d",&num[i]);
        l[1]=1;
        REPF(i,2,n)
        {
            int pos=i;
            while(pos-1>=1&&num[i]<=num[pos-1])
                pos=l[pos-1];
            l[i]=pos;
        }
        r[n]=n;
        for(int i=n-1;i>=1;i--)
        {
            int pos=i;
            while(pos+1<=n&&num[i]<=num[pos+1])
                pos=r[pos+1];
            r[i]=pos;
        }
        LL ans=0;
        REPF(i,1,n)
            ans=max(ans,(LL)num[i]*(r[i]-l[i]+1));
        printf("%lld\n",ans);
    }
    return 0;
}