POJ 2559 Largest Rectangle in a Histogram(单调栈)

【题目链接】:click here~~

【题目大意】:

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.


Sample Input

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

Sample Output

8
4000
【解题思路】:
建立一个单调递增栈,所有元素各进栈和出栈一次,每次出栈的时候更新一下最大值

代码:

// C
#ifndef _GLIBCXX_NO_ASSERT
#include 
#endif

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using namespace std;
#define rep(i,j,k) for(int i=j;i(int)k;--i)

#define lowbit(a) a&-a
#define Max(a,b) a>b?a:b
#define Min(a,b) a>b?b:a
#define mem(a,b) memset(a,b,sizeof(a))
typedef long long LL;
typedef unsigned long long LLU;
typedef double db;

const int N=2*1e5+10;
int n,m,top;
int  num[N],W[N],H[N];

char str[N];
bool vis[N];

int dir4[4][2]= {{1,0},{0,1},{-1,0},{0,-1}};
int dir8[8][2]= {{1,0},{1,1},{0,1},{-1,1},{-1,0},{-1,-1},{0,-1},{1,-1}};

struct node      ///定义栈的结构体,高度:宽度
{
    int height;
    int width;
};
node Dull_Stack[N];

int main()
{
    LL ans,tot,tmp,Max_area;
    while(scanf("%d",&n)!=EOF&&n)
    {
        ans=0;
        top=0;
        rep(i,0,n)
        {
            scanf("%d",&m);
            tmp=0;
            while(top>0&&Dull_Stack[top-1].height>=m)///(2,1) (1,2)的情况
            {
                tot=Dull_Stack[top-1].height*(Dull_Stack[top-1].width+tmp);///更新宽度
                //ans=max(tot,ans);
                if(tot>ans) ans=tot;
                tmp+=Dull_Stack[top-1].width;
                --top;
                /*
                printf("Dull_Stack[top-1].height= %lld\n",Dull_Stack[top-1].height);
                printf("Dull_Stack[top-1].width= %lld\n",Dull_Stack[top-1].width);
                printf("ans= %lld\n",ans);
                printf("tot= %lld\n",tot);
                */
            }
            Dull_Stack[top].height=m;///继续输入
            Dull_Stack[top].width=tmp+1;
            ++top;
        }
        /*
         printf("tot=%lld\n",tot);
          printf("ans=%lld\n",ans);
          */
        tmp=0;
        while(top>0)
        {
            Max_area=Dull_Stack[top-1].height*(Dull_Stack[top-1].width+tmp);
            //ans=max(Max_area,ans);
            if(Max_area>ans) ans=Max_area;
            tmp+=Dull_Stack[top-1].width;
            ///printf("%lld\n",Max_area);
            --top;
        }
        printf("%lld\n",ans);
    }
    return 0;
}
/*
Sample Input
7
2 1 4 5 1 3 3
4
1000 1000 1000 1000
0
Sample Output
8
4000
*/





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