HDU1506-Largest Rectangle in a Histogram（dp）

题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=1506

思路

就只需要找到当前每一个矩形，以它的高度，向左向右分别能延伸多长即可
直接暴力肯定会T，其实每次能延伸的都和上一次有关系，因此考虑dp
状态表示

l[i]：以当前矩形的高度向左能延伸多少

转移
如果h[i] ≥ h[i - 1]，那么l[i] = 1，但是在i - 1 - l[i - 1]之前，还存在高度在h[i - 1]和h[i]之间的，还需要累加上

代码

#include <iostream>#include <cstring>#include <stack>#include <vector>#include <set>#include <map>#include <cmath>#include <queue>#include <sstream>#include <iomanip>#include <fstream>#include <cstdio>#include <cstdlib>#include <climits>#include <deque>#include <bitset>#include <algorithm>using namespace std;#define PI acos(-1.0)#define LL long long#define PII pair<int, int>#define PLL pair<LL, LL>#define mp make_pair#define IN freopen("in.txt", "r", stdin)#define OUT freopen("out.txt", "wb", stdout)#define scan(x) scanf("%d", &x)#define scan2(x, y) scanf("%d%d", &x, &y)#define scan3(x, y, z) scanf("%d%d%d", &x, &y, &z)#define sqr(x) (x) * (x)#define pr(x) cout << #x << " = " << x << endl#define lc o << 1#define rc o << 1 | 1#define pl() cout << endl#define CLR(a, x) memset(a, x, sizeof(a))#define FILL(a, n, x) for (int i = 0; i < n; i++) a[i] = xconst int maxn = 100000 + 5;const int INF = 0x3e3e3e3e;int l[maxn], r[maxn], a[maxn], n;LL solve() {    LL res = 0;    l[0] = r[n + 1] = 0;    a[0] = a[n + 1] = INF;    for (int i = 1; i <= n; i++) {        l[i] = 1;        int t = i - 1;        while (a[i] <= a[t] && t >= 1) {            l[i] += l[t];            t -= l[t];        }    }    for (int i = n; i >= 1; i--) {        r[i] = 1;        int t = i + 1;        while (a[i] <= a[t] && t <= n) {            r[i] += r[t];            t += r[t];        }    }    for (int i = 1; i <= n; i++) {        LL t = (LL)a[i] * (LL)(l[i] + r[i] - 1);        res = max(res, t);    }    return res;}int main() {    while (scan(n) == 1 && n) {        for (int i = 1; i <= n; i++) scan(a[i]);        printf("%I64d\n", solve());    }    return 0;}