stack(单调栈) POJ 2082 Terrible Sets

题目传送门

题意：紧贴x轴有一些挨着的矩形，给出每个矩形的长宽，问能组成的最大矩形面积为多少

分析：用堆栈来维护高度递增的矩形，遇到高度小的，弹出顶部矩形直到符合递增，顺便计算矩形面积，且将弹出的宽度都累积到当前的矩形中，这样最后再扫描一遍，算面积很方便，这题应该算是 POJ 2559 的强化版了

收获：stack的应用，求矩形面积，矩阵相乘，表达式计算

代码：

/************************************************
* Author        :Running_Time
* Created Time  :2015/9/9 星期三 13:50:48
* File Name     :L.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 5e4 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
struct M    {
    int w, h;
}m[N];

int main(void)    {
    int n;
    while (scanf ("%d", &n) == 1)   {
        if (n == -1)    break;
        for (int i=1; i<=n; ++i)    {
            scanf ("%d%d", &m[i].w, &m[i].h);
        }
        stack<M> S;
        int ans = 0, lasth = 0;
        for (int i=1; i<=n; ++i)    {
            if (m[i].h >= lasth)    {
                S.push (m[i]);  lasth = m[i].h;
            }
            else    {
                int totw = 0, area = 0;
                while (!S.empty () && S.top ().h > m[i].h)  {
                    totw += S.top ().w;
                    area = totw * S.top ().h;
                    if (area >= ans)    ans = area;
                    S.pop ();
                }
                m[i].w += totw;
                S.push (m[i]);  lasth = m[i].h;
            }
        }
        int totw = 0, area = 0;
        while (!S.empty ()) {
            totw += S.top ().w;
            area = totw * S.top ().h;
            if (area >= ans)    ans = area;
            S.pop ();
        }
        printf ("%d
", ans);
    }

    return 0;
}