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\n", ans);
}
return 0;
}
stack(单调栈) POJ 2082 Terrible Sets
原文：http://www.cnblogs.com/Running-Time/p/4795754.html