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POJ2082---Terrible Sets(單調棧)

編輯：C++入門知識

POJ2082---Terrible Sets(單調棧)

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+…+wnhn < 109.

Output
Simply output Max(S) in a single line for each case.

Sample Input

3
1 2
3 4
1 2
3
3 4
1 2
3 4
-1

Sample Output

12
14

Source

一道水題,題面那麼難懂…
就是給你一排矩形,求可以形成的最大的矩形面積是多少
單調棧搞定

/*************************************************************************
    > File Name: POJ2082.cpp
    > Author: ALex
    > Mail: [email protected] 
    > Created Time: 2015年05月07日 星期四 20時11分08秒
 ************************************************************************/

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include
#include 
#include 
#include 

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair  PLL;

static const int N = 50010;
int w[N], h[N];
int r[N], l[N];
int sum[N];
stack  st;

int main() {
    int n;
    while (~scanf("%d", &n) && n != -1) {
        for (int i = 1; i <= n; ++i) {
            scanf("%d%d", &w[i], &h[i]);
            l[i] = r[i] = i;
        }
        while (!st.empty()) {
            st.pop();
        }
        sum[0] = 0;
        for (int i = 1; i <= n; ++i) {
            sum[i] = sum[i - 1] + w[i];
        }
        for (int i = n; i >= 1; --i) {
            if (st.empty()) {
                st.push(make_pair(h[i], i));
            }
            else {
                while (!st.empty()) {
                    PLL u = st.top();
                    if (u.first <= h[i]) {
                        break;
                    }
                    st.pop();
                    l[u.second] = i + 1;
                }
                st.push(make_pair(h[i], i));
            }
        }
        while (!st.empty()) {
            PLL u = st.top();
            st.pop();
            l[u.second] = 1;
        }
        for (int i = 1; i <= n; ++i) {
            if (st.empty()) {
                st.push(make_pair(h[i], i));
            }
            else {
                while (!st.empty()) {
                    PLL u = st.top();
                    if (u.first <= h[i]) {
                        break;
                    }
                    st.pop();
                    r[u.second] = i - 1;
                }
                st.push(make_pair(h[i], i));
            }
        }
        while (!st.empty()) {
            PLL u = st.top();
            st.pop();
            r[u.second] = n;
        }
        int ans = 0;
        for (int i = 1; i <= n; ++i) {
            int L = l[i];
            int R = r[i];
            ans = max(ans, h[i] * (sum[R] - sum[L - 1]));
        }
        printf("%d\n", ans);
    }
    return 0;
}