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, ... , AN (1 ≤ N ≤ 2,000) describing the elevation (0 ≤ Ai ≤ 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, . ... , BN 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| + ... + |AN - BN |
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: Ai
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
对于这道题我刚开始的想法是DP[前i个数][最后一个数的大小]，后来发现其数据范围太大 a的范围为1e9
肯定不行，后来又看到了离散化，把n个数的值作为二维值这样时间复杂度就降下来了。
状态转移方程：dp[i-1][j]=ABS(a[i]-b[j])+min(dp[i-1][j])；
#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<queue>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
using namespace std;
typedef long long ll;
const int INF_INT = 0x3f3f3f3f;
const ll INF_LL = 1e18;
const int MAXN =1e5+5;
const int MAXM = 1015;
ll dp[3005][3015];
int n;
ll a[3005];
ll b[3005];
int main()
{
// ios_base::sync_with_stdio(false);
// freopen("data.txt","r",stdin);
while(cin>>n)
{
for(int i=1;i<=n;i++)
{
cin >> a[i];
b[i] = a[i];
}
sort(b+1,b+n+1);
ll minn;
for(int i=1;i<=n;i++)
{
minn = INF_LL;
for(int j=1;j<=n;j++)
{
minn = min(dp[i-1][j],minn);
dp[i][j] = minn+abs(a[i]-b[j]);
}
}
ll ans = INF_LL;
// for(int i=1;i<=n;i++)
// {
// for(int j=1;j<=n;j++)
// {
// cout<<dp[i][j]<<" ";
// }
// cout <<endl;
// }
for(int i=1;i<=n;i++)
{
ans = min(ans,dp[n][i]);
}
cout <<ans<<endl;
}
}
另一道Codeforces 713C C. Sonya and Problem Wihtout a Legend与之类似
而本题的严格如何转化非严格，直接加一行代码，a[i]-=i；
原理
推导过程大致如下：a[i] < a[i+1] —> a[i] <= a[i+1]-1 —> a[i]-i <= a[i+1]-(i+1)—> b[i]<=b[i+1]
#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<queue>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<map>
using namespace std;
typedef long long ll;
const int INF_INT = 0x3f3f3f3f;
const ll INF_LL = 1e18;
const int MAXN =1e5+5;
const int MAXM = 1015;
ll dp[3005][3015];
int n;
ll a[3005];
ll b[3005];
int main()
{
// ios_base::sync_with_stdio(false);
// freopen("data.txt","r",stdin);
while(cin>>n)
{
for(int i=1;i<=n;i++)
{
cin >> a[i];
a[i] = b[i] = a[i]-i;
}
sort(b+1,b+n+1);
ll minn;
for(int i=1;i<=n;i++)
{
minn = INF_LL;
for(int j=1;j<=n;j++)
{
minn = min(dp[i-1][j],minn);
dp[i][j] = minn+abs(a[i]-b[j]);
}
}
ll ans = INF_LL;
// for(int i=1;i<=n;i++)
// {
// for(int j=1;j<=n;j++)
// {
// cout<<dp[i][j]<<" ";
// }
// cout <<endl;
// }
for(int i=1;i<=n;i++)
{
ans = min(ans,dp[n][i]);
}
cout <<ans<<endl;
}
}