POJ 3666 Making the Grade (动态规划)

佚名 6年前 (2019-04-08) 随笔 1854人围观抢沙发百度已收录

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).

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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

| A ₁- B ₁| + | A ₂- B ₂| + ... + | 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

Source

USACO 2008 February Gold 题意： 将所给数组中的某个数字加上或者减去某个数，使数组变为非降数组，问所需最小花费。 思路： 允许数组中的数字相等，那么最后最优解不会出现除了输入以外的数字,所以可以将输入的数字离散化。 dp[i][j]表示，将第i个数字,变成第j大的数字所需的最小花费。j实际上就是离散化之后的数组的下标。 dp[i][j]=min(dp[i-1][k]+abs(num[i]-p[k]),dp[i][j]); 其中num是原高度，p是离散化后的数组。k<=j; 但是这样的复杂度是n的三次方，不过还好我们可以用一个数组记录下j之前dp[i-1][k]+abs(num[i]-p[k])的最小值，这样就能优化成n方了。 TLE

#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#define fuck(x) cout<<#x<<" = "<<x<<endl;
#define debug(x,i,j) cout<<#x<<"["<<i<<"]["<<j<<"] = "<<x[i][j]<<endl;
#define ls (t<<1)
#define rs ((t<<1)+1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 100086;
const int inf = 2.1e9;
const ll Inf = 999999999999999999;
const int mod = 1000000007;
const double eps = 1e-6;
const double pi = acos(-1);
int num[2008],p[2008];
int n;
int dp[2008][2008];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d",&num[i]);
        p[i]=num[i];
    }
    sort(p+1,p+1+n);
    int m=unique(p+1,p+1+n)-p-1;
    for(int i=1;i<=n;i++){
        int t=lower_bound(p+1,p+1+n,num[i])-p-1;
        for(int j=1;j<=m;j++){
            dp[i][j]=inf;
            for(int k=1;k<=j;k++){
                dp[i][j]=min(dp[i-1][k]+abs(num[i]-p[k]),dp[i][j]);
            }
        }
    }
    printf("%d\n",dp[n][m]);
    return 0;
}

View Code AC

#include<iostream>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#define fuck(x) cout<<#x<<" = "<<x<<endl;
#define debug(x,i,j) cout<<#x<<"["<<i<<"]["<<j<<"] = "<<x[i][j]<<endl;
#define ls (t<<1)
#define rs ((t<<1)+1)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 100086;
const int inf = 2.1e9;
const ll Inf = 999999999999999999;
const int mod = 1000000007;
const double eps = 1e-6;
const double pi = acos(-1);
int num[2008],p[2008];
int n;
int dp[2008][2008];
int minn[2008];
int main()
{
//    ios::sync_with_stdio(false);
//    freopen("in.txt","r",stdin);
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d",&num[i]);
        p[i]=num[i];
    }
    sort(p+1,p+1+n);
    int m=unique(p+1,p+1+n)-p-1;
    for(int i=1;i<=n;i++){
        int t=lower_bound(p+1,p+1+n,num[i])-p-1;
        minn[0]=inf;
        for(int j=1;j<=m;j++){
            minn[j]=min(minn[j-1],dp[i-1][j]+abs(num[i]-p[j]));
        }
        for(int j=1;j<=m;j++){
            dp[i][j]=minn[j];
        }
    }
    printf("%d\n",dp[n][m]);
    return 0;
}