Terrible Sets

Time Limit: 1000MS	Memory Limit: 30000K
Total Submissions: 2778	Accepted: 1413

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

3
1 2
3 4
1 2
3
3 4
1 2
3 4
-1

Sample Output

12
14

Source

題意就是給你一個排成一排的小矩形，要你找出他們組成的最大的矩形。用單調棧預處理出每個矩形往左能到達的高度大於它的最遠點，和往右的最遠點即可。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<vector>
#include<cstring>
#include<queue>
#include<stack>
using namespace std;
const int maxn = 50000 + 5;
const int INF = 2000000000;
typedef  pair<int, int> P;
typedef long long LL;

int w[maxn],h[maxn];
int l[maxn],r[maxn];
int sum[maxn];
stack<P> S;

int main(){
    int n;
    while(scanf("%d",&n)){
        if(n == -1) break;
        for(int i = 0;i < n;i++){
            scanf("%d%d",&w[i],&h[i]);
        }
        sum[0] = w[0];
        for(int i = 1;i < n;i++) sum[i] = sum[i-1]+w[i];
        while(!S.empty()) S.pop();
        for(int i = 0;i < n;i++){
            while(!S.empty() && h[i] <= S.top().first) S.pop();
            if(S.size() == 0) l[i] = sum[i];
            else l[i] = sum[i]-sum[S.top().second];
            S.push(P(h[i],i));
        }
        while(!S.empty()) S.pop();
        for(int i = n-1;i >= 0;i--){
            while(!S.empty() && h[i] <= S.top().first) S.pop();
            if(S.size() == 0) r[i] = sum[n-1]-sum[i-1];
            else r[i] = sum[S.top().second-1]-sum[i-1];
            S.push(P(h[i],i));
        }
        int ans = 0;
        for(int i = 0;i < n;i++){
            ans = max(ans,h[i]*(l[i]+r[i]-w[i]));
        }
        printf("%d\n",ans);
    }
    return 0;
}