[dp]POJ2559 && HDOJ1506 Largest Rectangle in a Histogram

题意 给n个条形的高度, 问能放的最大矩形面积

分析: 从左到右 从右到左 各搞一遍

分别记录      L[i]记录列(从前往后)标 第几列开始 可以往后放高度为a[i]的矩形

       R[i]记录列(从后往前)标 第几列开始 可以往前放高度为a[i]的矩形

 R[i]-L[i]+1即为高度为a[i]的矩形能横穿的列数

再乘个a[i]即为面积  遍历所有高度 求最大值即可

ps.注意范围 面积要用LL

pps.注意 max((int), (LL)) 的CE

 1 #include <cstdio>
 2 #include <cstdlib>
 3 #include <cstring>
 4 #include <climits>
 5 #include <cctype>
 6 #include <cmath>
 7 #include <string>
 8 #include <sstream>
 9 #include <iostream>
10 #include <algorithm>
11 #include <iomanip>
12 using namespace std;
13 #include <queue>
14 #include <stack>
15 #include <vector>
16 #include <deque>
17 #include <set>
18 #include <map>
19 typedef long long LL;
20 typedef long double LD;
21 #define pi acos(-1.0)
22 #define lson l, m, rt<<1
23 #define rson m+1, r, rt<<1|1
24 typedef pair<int, int> PI;
25 typedef pair<int, PI> PP;
26 #ifdef _WIN32
27 #define LLD "%I64d"
28 #else
29 #define LLD "%lld"
30 #endif
31 //#pragma comment(linker, "/STACK:1024000000,1024000000")
32 //LL quick(LL a, LL b){LL ans=1;while(b){if(b & 1)ans*=a;a=a*a;b>>=1;}return ans;}
33 //inline int read(){char ch=' ';int ans=0;while(ch<'0' || ch>'9')ch=getchar();while(ch<='9' && ch>='0'){ans=ans*10+ch-'0';ch=getchar();}return ans;}
34 inline void print(LL x){printf(LLD, x);puts("");}
35 //inline void read(double &x){char c = getchar();while(c < '0') c = getchar();x = c - '0'; c = getchar();while(c >= '0'){x = x * 10 + (c - '0'); c = getchar();}}
36 
37 #define N 100005
38 int a[N], L[N], R[N];
39 int main()
40 {
41 #ifndef ONLINE_JUDGE
42     freopen("in.txt", "r", stdin);
43     freopen("out.txt", "w", stdout);
44 #endif
45     int n;
46     while(~scanf("%d", &n) && n)
47     {
48         for(int i=0;i<n;i++)
49             scanf("%d", &a[i]);
50         memset(L, 0, sizeof(L));
51         memset(R, 0, sizeof(R));
52         for(int i=0;i<n;i++)
53         {
54             L[i]=i;
55             while(L[i]>0 && a[L[i]-1]>=a[i])
56                 L[i]=L[L[i]-1];
57         }
58         for(int i=n-1;i>=0;i--)
59         {
60             R[i]=i;
61             while(R[i]<n-1 && a[R[i]+1]>=a[i])
62                 R[i]=R[R[i]+1];
63         }
64         LL ans=0;
65         for(int i=0;i<n;i++)
66             ans=max(ans, (LL)(R[i]-L[i]+1)*a[i]);
67         print(ans);
68     }
69     return 0;
70 }
HDOJ1506
原文地址:https://www.cnblogs.com/Empress/p/4082778.html