题意：
给你n 个宽度为1 高度不同的矩形，要求求最大的矩形面积， 高度低的可以向挨着的高度高的扩展。
思路：
很多都是单调栈，dp之类的。
给大家 提供另一个思路：
枚举矩形中心，二分右边最远能到哪，二分左边最远能到哪，更新最大值即可。
一个区间合不合适，只要这个区间的最小值是否等于中心的高度即可。可以用RMQ处理。
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn = 100000 + 7;
int MIN[maxn<<2];
int a[100000+7];
const int MAXN = maxn;
int dp[MAXN][20];
int mm[MAXN];
void initRMQ(int n,int b[])
{
mm[0] = -1;
for(int i = 1; i <= n;i++)
{
mm[i] = ((i&(i-1)) == 0)?mm[i-1]+1:mm[i-1];
dp[i][0] = b[i];
}
for(int j = 1; j <= mm[n];j++)
for(int i = 1;i + (1<<j) -1 <= n;i++)
dp[i][j] = min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
}
int rmq(int x,int y)
{
int k = mm[y-x+1];
return min(dp[x][k],dp[y-(1<<k)+1][k]);
}
int main(){
int n;
while(~scanf("%d",&n) && n){
for (int i = 1; i <= n; ++i){
scanf("%d",a+i);
}
initRMQ(n,a);
long long ans = -1;
for (int i = 1; i <= n; ++i){
int L = i,R = n;
int ans1=i,ans2=i;
while(L <= R){
int m = L+R>>1;
if (m>=i && rmq(i,m) == a[i]){
ans2 = m;
L=m+1;
}
else R=m-1;
}
L = 1;
R = i;
while(L <=R){
int m = L+R>>1;
if (m <=i && rmq(m,i) == a[i]){
ans1 = m;
R = m-1;
}
else L = m+1;
}
long long t = (long long)(ans2-ans1+1)*(long long )a[i] ;
if (ans < t) ans =t;
}
printf("%lld\n",ans);
}
return 0;
}
/**
4
4 4
5 2
11 5
15 10
25 100
**/Largest Rectangle in a Histogram Description A histogram is a polygon composed of a sequence of rectangles aligned at a common base line. The rectangles have equal widths but may have different heights. For example, the figure on the left shows the histogram that consists of rectangles with the heights 2, 1, 4, 5, 1, 3, 3, measured in units where 1 is the width of the rectangles:
Usually, histograms are used to represent discrete distributions, e.g., the frequencies of characters in texts. Note that the order of the rectangles, i.e., their heights, is important. Calculate the area of the largest rectangle in a histogram that is aligned at the common base line, too. The figure on the right shows the largest aligned rectangle for the depicted histogram. Input The input contains several test cases. Each test case describes a histogram and starts with an integer n, denoting the number of rectangles it is composed of. You may assume that 1<=n<=100000. Then follow n integers h1,...,hn, where 0<=hi<=1000000000. These numbers denote the heights of the rectangles of the histogram in left-to-right order. The width of each rectangle is 1. A zero follows the input for the last test case. Output For each test case output on a single line the area of the largest rectangle in the specified histogram. Remember that this rectangle must be aligned at the common base line. Sample Input 7 2 1 4 5 1 3 34 1000 1000 1000 10000 Sample Output 84000 Hint Huge input, scanf is recommended. Source |
Time Limit: 1000MS | | Memory Limit: 65536K |
Total Submissions: 21013 | | Accepted: 6754 |
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