POJ - 3666 Making the Grade

最新推荐文章于 2022-05-09 16:57:25 发布

Lngxling

最新推荐文章于 2022-05-09 16:57:25 发布

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分类专栏：思维队列&&优先队列

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本文介绍了一个经典的算法问题——如何通过最小的操作次数使一条由多个高度点组成的道路变为单调递增或递减的状态。该问题通过一种高效的方法解决，具体思路是在遇到不符合单调性的高度时，选择减少前面的高度来保持整体成本最低。

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Making the Grade

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8186		Accepted: 3831

Description

A straight dirt road connects two fields on FJ's farm, but it changes elevation more than FJ would like. His cows do not mind climbing up or down a single slope, but they are not fond of an alternating succession of hills and valleys. FJ would like to add and remove dirt from the road so that it becomes one monotonic slope (either sloping up or down).

You are given N integers A₁, ... , A_N (1 ≤ N ≤ 2,000) describing the elevation (0 ≤ A_i ≤ 1,000,000,000) at each of N equally-spaced positions along the road, starting at the first field and ending at the other. FJ would like to adjust these elevations to a new sequence B₁, . ... , B_N that is either nonincreasing or nondecreasing. Since it costs the same amount of money to add or remove dirt at any position along the road, the total cost of modifying the road is

|A₁- B₁| + |A₂- B₂| + ... + |A_N - B_N |

Please compute the minimum cost of grading his road so it becomes a continuous slope. FJ happily informs you that signed 32-bit integers can certainly be used to compute the answer.

Input

* Line 1: A single integer: N
* Lines 2..N+1: Line i+1 contains a single integer elevation: A_i

Output

* Line 1: A single integer that is the minimum cost for FJ to grade his dirt road so it becomes nonincreasing or nondecreasing in elevation.

Sample Input

Sample Output

题意：给定n个数的序列，可以执行一个操作，将某一个数加一或者减一，要使得这个序列递增，问最少的操作次数。

思路：当遇到一个数比前面的数小时，我们可以把它加到前面的数，也可以把前面的数减到当前的数，两者所需的操作次数是相等的。我们选择了将前面的数减小，这样就可以让后面输入的数可以小一些。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <stack>
#include <map>
#include <cmath>
#include <vector>
#define max_ 200010
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;

int n;
ll ans;
priority_queue<int>q;
int main(int argc, char const *argv[]) {
    cin>>n;
    ans=0;
    while(n--)
    {
        int x;
        cin>>x;
        q.push(x);
        if(x<q.top())
        {
            ans+=(q.top()-x);
            q.pop();
            q.push(x);
        }
    }
    printf("%lld\n",ans);
    return 0;
}