POJ - 2559 Largest Rectangle in a Histogram(单调栈)

题意:有n个高度不同的直方图,求直方图内最大的矩形面积。

POJ - 2559 Largest Rectangle in a Histogram(单调栈)_第1张图片

分析:

1、若当前研究高度大于栈顶高度,则直接入栈。否则,边处理栈内所有高度大于等于当前高度的元素边出栈,在此过程中,边累加宽度边以当前栈顶元素为高算出矩形面积,比较最大值,直到最终将比当前高度大的元素都捋平,将捋平后的高度即当前高度,和最终累积的宽度入栈。

2、上述处理的结果,使得栈内所剩的元素都是从栈顶到栈底高度递减,按照与1相同的处理方法计算栈内元素的矩形面积,比较最大值即可。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 100000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
struct Node{
    int h, w;
    Node(){}
    Node(int hh, int ww):h(hh), w(ww){}
}num[MAXN];
stack s;
int main(){
    int n;
    while(scanf("%d", &n) == 1){
        if(!n) return 0;
        for(int i = 0; i < n; ++i){
            scanf("%d", &num[i].h);
        }
        LL ans = 0;
        for(int i = 0; i < n; ++i){
            int width = 0;
            while(!s.empty() && s.top().h >= num[i].h){
                int tmph = s.top().h;
                int tmpw = s.top().w;
                s.pop();
                width += tmpw;
                ans = Max(ans, (LL)tmph * width);
            }
            s.push(Node(num[i].h, width + 1));
        }
        int t = 0;
        while(!s.empty()){
            Node tmp = s.top();
            s.pop();
            ans = Max(ans, (LL)tmp.h * (t + tmp.w));
            t += tmp.w;
        }
        printf("%lld\n", ans);
    }
    return 0;
}

  

转载于:https://www.cnblogs.com/tyty-Somnuspoppy/p/6547291.html

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