ZOJ 2422 POJ 2082 Terrible Sets

Terrible Sets

Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 5104 Accepted: 2611
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+…+wnhn < 109.
Output
Simply output Max(S) in a single line for each case.
Sample Input
3
1 2
3 4
1 2
3
3 4
1 2
3 4
-1
Sample Output
12
14
Source
Shanghai 2004 Preliminary

维护一个矩形的高度单调递增的栈。。。

这个人的讲解不错的，图画的很用心：传送门
ZOJ上不去了 没在上面交。。。may be会WA，到时候勿喷！

#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>#include<stack>using namespace std; struct Rect{    int w, h;    Rect (int w, int h) : w(w), h(h){}};inline void read(int &x){    x=0; int f=1; char c=getchar();    while(c>'9'||c<'0'){ if(c=='-')f=-1; c=getchar(); }    while(c>='0'&&c<='9'){ x=x*10+c-'0'; c=getchar(); } x*=f;}int main(int argc,char *argv[]){    int n;    while(1){        read(n);        if(n == -1) break;        stack<Rect> s; int Ans = 0;        for(int w,h,i=1; i<=n; ++i){            read(w),read(h);            if(!s.empty()){                if(h >= s.top().h) s.push(Rect(w, h));                else {                    int tot_w = 0;                    while(!s.empty()) {                        if(h > s.top().h) break;                        tot_w += s.top().w;                        Ans = max(Ans , tot_w * s.top().h);                        s.pop();                    }                    s.push(Rect(tot_w + w,h));                }            } else s.push(Rect(w,h));        }        int tot_w = 0;        while(!s.empty()){            tot_w += s.top().w;            Ans = max(Ans , tot_w * s.top().h);            s.pop();        }        printf("%d\n", Ans);    }    return 0;}