POJ 3666 Making the Grade

Making the Grade

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3425		Accepted: 1601

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

7
1
3
2
4
5
3
9

Sample Output

3

Source

USACO 2008 February Gold

一开始想到的是贪心

从头开始不下降

从头开始不上升

从尾开始不下降

从尾开始不上升

这样每一次不是全部砍掉就是全部向上建造

最后得分40分

讲解完后了解到这是个dp题

先算不下降的

f[i][j]表示第i个高度j时的最小花费

丧心病狂的数据 0 ≤ j ≤ 1,000,000,000

但是发现最多只有2,000个建筑

即最多只有2000个不同的高度

可以枚举出现的这些高度 因为调整后的高度一定是出现过的高度 这样才是最优的

可以先读入每一个高度h[i] 排序后存入t中

f[i][j]=min(f[i-1][k])+abs(h[i]-t[j])

k<=j

这样是n^3的dp 需要循环i j k 可以得到50分

可以用g[i][j]维护min(f[i-1][k])

这样可以降低到n^2 可以过了

g[i][1]=f[i][1]

g[i][k]=min(g[i][k-1],f[i][k]),k>1

然后可以倒着做不下降 也可以正着做不上升 求最小值即为答案

//N^3
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;

int h[2022],t[2022];
int g[2022][2022],f[2022][2022];

int m,n,a,b,c,z;

int uplis(int i,int tall)
{
    if(i==0||tall<0)return 0;
    if(f[i][tall]!=-1)return f[i][tall];
    f[i][tall]=uplis(i-1,1)+abs(t[tall]-h[i]);
    for(int a=2;a<=tall;a++)
    f[i][tall]=min(f[i][tall],uplis(i-1,a)+abs(t[tall]-h[i]));
    return f[i][tall];
}

int downlis(int i,int tall)
{
    if(i==0||tall<0)return 0;
    if(f[i][tall]!=-1)return f[i][tall];
    f[i][tall]=downlis(i-1,m)+abs(t[tall]-h[i]);
    for(int a=m-1;a>=tall;a--)
    f[i][tall]=min(f[i][tall],downlis(i-1,a)+abs(t[tall]-h[i]));
    return f[i][tall];
}

int main()
{
    scanf("%d",&m);
    for(a=1;a<=m;a++){scanf("%d",&h[a]);t[a]=h[a];}
    sort(t+1,t+m+1);
    
    memset(f,-1,sizeof(f));
    memset(g,-1,sizeof(f));
    z=uplis(m,1);
    for(a=2;a<=m;a++)z=min(z,uplis(m,a));
    
    memset(f,-1,sizeof(f));
    memset(g,-1,sizeof(f));
    for(a=1;a<=m;a++)z=min(z,downlis(m,a));
    
    cout<<z<<'\n';
    return 0;
}

//N^2
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;

int h[2022],t[2022];
int g[2022][2022],f[2022][2022];

int m,n,a,b,c,z=999999999;

void uplis()
{
    memset(f,0,sizeof(f));
    memset(g,0,sizeof(g));
    int i,j;
    for(i=1;i<=m;++i)
        for(j=1;j<=m;++j)
        {
            f[i][j]=g[i-1][j]+abs(h[i]-t[j]);
            if(j==1)g[i][j]=f[i][j];
            else g[i][j]=min(f[i][j],g[i][j-1]);
        }
    for(int a=1;a<=m;a++)z=min(z,f[m][a]);
}

void downlis()
{
    memset(f,0,sizeof(f));
    memset(g,0,sizeof(g));
    int i,j;
    for(i=1;i<=m;++i)
        for(j=m;j>=1;--j)
        {
            f[i][j]=g[i-1][j]+abs(h[i]-t[j]);
            if(j==m)g[i][j]=f[i][j];
            else g[i][j]=min(f[i][j],g[i][j+1]);
        }
    for(int a=1;a<=m;a++)z=min(z,f[m][a]);
}

int main()
{
    scanf("%d",&m);
    for(a=1;a<=m;a++){scanf("%d",&h[a]);t[a]=h[a];}
    sort(t+1,t+m+1);
    uplis();
    downlis();
    cout<<z<<'\n';
    return 0;
}