POJ 2082 Terrible Sets（可怕的矩形）

。。一个好题。。

Terrible Sets

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 4812 Accepted: 2465

Description

Let N be the set of all natural numbers {0 , 1 , 2 , . . . }, and R be the set of all real numbers. wi, hi for i = 1 . . . n are some elements in N, and w0 = 0. 
Define set B = {< x, y > | x, y ∈ R and there exists an index i > 0 such that 0 <= y <= hi ,∑_0<=j<=i-1wj <= x <= ∑_0<=j<=iwj} 
Again, define set S = {A| A = WH for some W , H ∈ R⁺ and there exists x0, y0 in N such that the set T = { < x , y > | x, y ∈ R and x0 <= x <= x0 +W and y0 <= y <= y0 + H} is contained in set B}. 
Your mission now. What is Max(S)? 
Wow, it looks like a terrible problem. Problems that appear to be terrible are sometimes actually easy. 
But for this one, believe me, it's difficult.

Input

The input consists of several test cases. For each case, n is given in a single line, and then followed by n lines, each containing wi and hi separated by a single space. The last line of the input is an single integer -1, indicating the end of input. You may assume that 1 <= n <= 50000 and w₁h₁+w₂h₂+...+w_nh_n < 10⁹.

Output

Simply output Max(S) in a single line for each case.

Sample Input

31 23 41 233 41 23 4-1

Sample Output

1214

Source

Shanghai 2004 Preliminary

题意：

在 x 轴上给定一些不会重叠的矩形宽和高，求解出可以形成的最大矩形的面积。

思路：

我想的用的贪心，明显的会 GG。思路就是用一个按照高度 h 非递减的方式将已知矩形进栈，如果即将入栈的高度 data.h 比栈顶元素的高度 lasth 小。

就一直退到可以保持栈顶元素高度非递减的那个元素 x 位置，退栈过程中也不能闲着，统计由 lasth 到 x 之间可以形成的最大矩形的面积 S ，以及之间矩形的

总宽度 totalw，弹出以上所有元素之后再压入一个高度为 data.h，宽度为 towalw + data.w 的矩形就可以啦，继续键入下一个矩形......最后扫描一遍。

样例1 - >

参考来源：《ACM / ICPC 算法训练教程（余立功）》

AC CODE：

#include<cstdio>#include<stack>using namespace std;struct rec {int w;int h;} data;int main(){int n, ans, i, lasth, totalw, curarea;while(scanf("%d", &n) && n != -1) {ans = 0;stack<rec> s;lasth = 0;//上次进栈的矩形高度for(i = 0; i < n; i++) {scanf("%d%d", &data.w, &data.h);if(data.h >= lasth) {s.push(data);} else {//当前进栈的矩形高度较小totalw = 0;//总宽curarea = 0;//当前面积while(!s.empty() && s.top().h > data.h) {totalw += s.top().w;curarea = totalw * s.top().h;if(curarea > ans) ans = curarea;s.pop();}totalw += data.w;data.w = totalw;s.push(data);//新的矩形 }lasth = data.h;}totalw = 0;curarea = 0;//当前面积 while(!s.empty()) {totalw += s.top().w;curarea = totalw * s.top().h;if(curarea > ans) ans = curarea;s.pop();}printf("%d\n", ans);}return 0;}/*41 32 33 34 33043 43 33 23 118*/