题干：
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 3
4 1000 1000 1000 1000
0
Sample Output
8
4000
解题报告：
单调栈模板！此题用动态规划亦可解。题目大意就是求矩形的可最大面积
ac代码1：（因为第一次用单调栈所以代码十分丑陋、、、）
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int MAX = 100000 + 5 ;
const int INF = 0x3f3f3f3f;
int top;
stack <ll> sk,sk2;
ll a[MAX];
ll zuo[MAX],you[MAX];
ll maxx;
int main()
{
int n;
while(scanf("%d",&n) ) {
if( n==0 ) break;
maxx=0;
while(!sk.empty() ) sk.pop();
while(!sk2.empty() ) sk2.pop();
top = -1;
for(int i = 0; i<n; i++) {
scanf("%lld",&a[i]);
}
//递增栈 找左侧的最小元素
sk.push(0);zuo[0]=-1;
for(int i = 1; i<n; i++) {
if(sk.empty() ) {
zuo[i]=-1;
sk.push(i);
}
else {
while(!sk.empty() && a[sk.top()]>=a[i] ) {
sk.pop();
}
if(sk.empty() ) {
zuo[i]=-1;
sk.push(i);
}
else {
zuo[i]=sk.top();
sk.push(i);
}
}
}
sk2.push(n-1);you[n-1]=n;
for(int i = n-2; i>=0; i--) {
if(sk2.empty() ) {
you[i]=n;
sk2.push(i);
}
else {
while(!sk2.empty() && a[sk2.top()]>=a[i] ) {
sk2.pop();
}
if(sk2.empty() ) {
you[i]=n;
sk2.push(i);
}
else {
you[i]=sk2.top();
sk2.push(i);
}
}
}
for(int i = 0; i<n; i++) {
maxx=max(maxx,a[i]*(you[i]-zuo[i]-1) );
}
printf("%lld\n",maxx);
}
return 0 ;
} ac代码2：（动态规划）
如果确定了长方形的左端点L和右端点R，那么最大可能的高度就是min{hi|L <= i < R}。
L[i] = (j <= i并且h[j-1] < h[i]的最大的j)
R[i] = (j > i并且h[j] > h[i]的最小的j)
#include <stdio.h>
#define MAX_N 100000
int n;
int h[MAX_N];
int L[MAX_N], R[MAX_N];
int stack[MAX_N];
long long max(long long a, long long b){
return (a > b) ? a : b;
}
void solve(){
//计算L
long long ans = 0;
int t = 0;
int i;
for (i = 0; i < n; ++i){
while (t > 0 && h[stack[t-1]] >= h[i]) t--;
L[i] = (t == 0) ? 0 : (stack[t-1] + 1);
stack[t++] = i;
}
//计算R
t = 0;
for (i = n - 1; i >= 0; --i){
while (t > 0 && h[stack[t-1]] >= h[i]) t--;
R[i] = (t == 0) ? n : stack[t-1];
stack[t++] = i;
}
for (i = 0; i < n; ++i){
ans = max(ans, (long long)h[i] * (R[i] - L[i]));
}
printf("%lld\n", ans);
}
int main(void){
int i;
while (scanf("%d", &n) != EOF && n != 0){
for (i = 0; i < n; ++i)
scanf("%d", &h[i]);
solve();
}
return 0;
} ac代码3：（动态规划）
/*
每一个图形都有一个h[i],l[i],r[i]！其中l[i]存的是左数至少比他高的图形的序数,r[i]存的是右数至少比他高的图形的序数！
剪枝 while(h[r[i]+1]>=h[i]) {r[i]=r[r[i]+1];}特别重要！看右面有没有比他大的，
有就看右面那个r[]存的序数是多少，一步一步最终找到连续的长方形！
*/
#include<cstdio>
#include<iostream>
const int N=110000;
using namespace std;
long long h[N],l[N],r[N];
int main()
{
long long n;
int i;
while(scanf("%lld",&n)&&n)
{
for(i=1;i<=n;i++)
{
scanf("%lld",&h[i]);
l[i]=r[i]=i;
}
h[0]=h[n+1]=-1;
for(i=1;i<=n;i++)
{
while(h[l[i]-1]>=h[i])
{
l[i]=l[l[i]-1];//这个体现动态规划的思想，存l[l[i]-1]而不是l[i]-1
}//这样就可以节省其中很多的步骤！根据动态规划原理这一找
}
for(i=n;i>=1;i--)
{
while(h[r[i]+1]>=h[i])
{
r[i]=r[r[i]+1];
}
}
long long maxs=-5,flag;
for(i=1;i<=n;i++)
{
flag=(r[i]-l[i]+1)*h[i];
if(flag>maxs)
{
maxs=flag;
}
}
printf("%lld\n",maxs);
}
} ac代码4：（单调栈模板）
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 100000 + 100;
stack<int> S;
ll h[N];
int R[N],L[N];
int main(){
int n;
while(~scanf("%d",&n) && n){
for(int i=0 ;i<n ;i++) scanf("%I64d",&h[i]);
while(S.size()) S.pop();
for(int i=0 ;i<n ;i++){
while(S.size() && h[S.top()] >= h[i]) S.pop();
if(S.empty()) L[i] = 0;
else L[i] = S.top() + 1;
S.push(i);
}
while(S.size()) S.pop();
for(int i=n-1 ;i>=0 ;i--){
while(S.size() && h[S.top()] >= h[i]) S.pop();
if(S.empty()) R[i] = n;
else R[i] = S.top();
S.push(i);
}
ll ans = 0;
for(int i=0 ;i<n ;i++){
ans = max(ans,h[i] * (R[i] - L[i]));
}
printf("%I64d\n",ans);
}
return 0;
}
总结：
1.单调栈的题 读数最好从1开始，不然就会出现ac代码4中S.top()+1这种不优雅的情况。
2.就算你是从0开始读的，但如果没有比L(i)小的，也要赋值坐标成0！因为你如果左边单调栈一次，右边单调栈一次，最后算差值的时候还要再-1（比如ac代码1），也会不美观。
3.综上所述，还是从i=1开始读比较好。。
4.注意数据是需要longlong的！！
5.下附单调栈模板（核心）：
for(int i=1 ;i<=n ;i++){
while(S.size() && h[S.top()] >= h[i]) S.pop();
if(S.empty()) L[i] = 0;
else L[i] = S.top();
S.push(i);
} 6.对上述模板有两个地方是可以更改的，一个是 h[S.top()] >= h[i] 的等于号，另一个是栈中存的数据，有时是存下标（位置），如此题，还有的情况是存值。这一点依题目情况而定。
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