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POJ2082---Terrible Sets(单调栈)

POJ2082---Terrible Sets(单调栈)

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

风流罗曼蒂克道水题,题面那么难懂…
纵使给您一排矩形,求能够变成的最大的矩形面积是微微
单调栈消除

/*************************************************************************
    > File Name: POJ2082.cpp
    > Author: ALex
    > Mail: [email protected] 
    > Created Time: 2015年05月07日 星期四 20时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("%dn", ans);
    }
    return 0;
}

Sets(单调栈) 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...

Time Limit: 1000MS   Memory Limit: 30000K
Total Submissions: 5067   Accepted: 2593

Terrible Sets
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.

Sample Output

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.

Shanghai 2004 Preliminary

Define set B = { | x, y R and there exists an index i > 0 such that 0 <= y <= hi, }

Terrible Sets

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

设想一个新的方框压到栈里面

Simply output Max in a single line for each case.

3、扫完后,把全路栈慢慢弹出,累积宽度更新答案

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.

POJ2082

Sample Input

这种东西很套路,正是用单调栈维护。

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 R and x0 <= x <= x0 W and y0 <= y <= y0 H} is contained in set B}.

 

Your mission now. What is Max?

 

12
14

1、一整个大方块

Sample Output

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