Terrible Sets--（单调队列）

Terrible Sets

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 60000/30000K (Java/Other)

Total Submission(s) : 29   Accepted Submission(s) : 16

Problem 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

单调队列常见题目，此题难在题意上题意读懂做就快了，与下面两个题目很像：

1.Largest Submatrix of All 1’s

2.Largest Rectangle in a Histogram

题目就是把以前宽度为1的柱子换为了宽度为w[i]的柱子

代码：

#include<iostream>#include<string>#include<cstdio>#include<algorithm>#include<cmath>#include<iomanip>#include<queue>#include<cstring>#include<map>using namespace std;typedef long long ll;#define pi acos(-1.0)#define inf 0x7fffffffint n,w[50005],h[50005],q[50005],l[50005],r[50005];ll sum[50005];void work(){    int i,j,tail=0;    for(i=1;i<=n;i++)    {        while(tail>=1&&h[q[tail]]>=h[i]) tail--;        if(tail==0) l[i]=1;        else l[i]=q[tail]+1;        q[++tail]=i;    }    tail=0;    for(i=n;i>=1;i--)    {        while(tail>=1&&h[q[tail]]>=h[i]) tail--;        if(tail==0) r[i]=n;        else r[i]=q[tail]-1;        q[++tail]=i;    }    ll cnt=0,temp=0;    for(i=1;i<=n;i++)    {        temp=(sum[r[i]]-sum[l[i]-1])*h[i];        if(temp>cnt) cnt=temp;    }    printf("%I64d\n",cnt);}int main(){    int i; sum[0]=0;    while(scanf("%d",&n)!=EOF)    {        if(n==-1) break;        for(i=1;i<=n;i++)        {            scanf("%d%d",&w[i],&h[i]);            sum[i]=sum[i-1]+w[i];            r[i]=l[i]=i;        }        work();    }    return 0;}