# 洛谷4267 [usaco18feb]taming the herd

## 考验预处理DP题

Posted by yjjr's blog on January 29, 2019

# 题目

## 题意翻译

Farmer John想要知道从他开始记录以来发生过多少次出逃。但是，他怀疑这些奶牛篡改了它的记录，现在他所确定的只有他是从发生出逃的某一天开始记录的。请帮助他求出，对于每个从他开始记录以来可能发生的出逃次数，他被篡改了的记录条数的最小值。

## 题目描述

Early in the morning, Farmer John woke up to the sound of splintering wood. It was the cows, and they were breaking out of the barn again! Farmer John was sick and tired of the cows’ morning breakouts, and he decided enough was enough: it was time to get tough. He nailed to the barn wall a counter tracking the number of days since the last breakout. So if a breakout occurred in the morning, the counter would be $0$ that day; if the most recent breakout was $3$ days ago, the counter would read $3$. Farmer John meticulously logged the counter every day.

The end of the year has come, and Farmer John is ready to do some accounting. The cows will pay, he says! But something about his log doesn’t look quite right…

Farmer John wants to find out how many breakouts have occurred since he started his log. However, he suspects that the cows have tampered with his log, and all he knows for sure is that he started his log on the day of a breakout. Please help him determine, for each number of breakouts that might have occurred since he started the log, the minimum number of log entries that must have been tampered with.

## 输入输出格式

### 输入格式

The first line contains a single integer $N$ ($1 \leq N \leq 100$), denoting the number of days since Farmer John started logging the cow breakout counter. The second line contains $N$ space-separated integers. The $i$th integer is a non-negative integer $a_i$ (at most $100$), indicating that on day $i$ the counter was at $a_i$, unless the cows tampered with that day’s log entry.

### 输出格式

The output should consist of $N$ integers, one per line. The $i$th integer should be the minimum over all possible breakout sequences with $i$ breakouts, of the number of log entries that are inconsistent with that sequence.

## 输入输出样例

### 输入样例#1

6
1 1 2 0 0 1


### 输出样例#1

4
2
1
2
3
4


## 说明

If there was only 1 breakout, then the correct log would look like 0 1 2 3 4 5, which is 4 entries different from the given log.

If there were 2 breakouts, then the correct log might look like 0 1 2 3 0 1, which is 2 entries different from the given log. In this case, the breakouts occurred on the first and fifth days.

If there were 3 breakouts, then the correct log might look like 0 1 2 0 0 1, which is just 1 entry different from the given log. In this case, the breakouts occurred on the first, fourth, and fifth days.

And so on.

Problem credits: Brian Dean and Dhruv Rohatgi

# 题解

## English

We can use Dynamic Programming to solve this task. Let $DP[i][j]$ means the minimum number of changes that during the first $i$ days there was $j$ breakouts.

Before the nomal Dynamic Programing, we should get $b[i][j]$ that means the minimum number of changes that if there was a breakout on day $i$, lasted until day $j$.

After that, the most important is State transition equation.

It means before day $i$, look for making a breakout on day $k$ that can let $DP[i][j]$ minimal.

The time Usage must be $O(N^3)$, because there are three Loop Variable $i,j,k$, and every Loop Variable will last from $1$ to $N$.

# Code

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define ll long long
#define mem(x,num) memset(x,num,sizeof x)
#define reg(x) for(int i=last[x];i;i=e[i].next)
using namespace std;
ll f=1,x=0;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
#define inf 0x3f3f3f
const int maxn=1e3+6;
int f[maxn][maxn],n,a[maxn],b[maxn][maxn];
int main(){
rep(i,1,n){
int tmp=0;
rep(j,i,n){
if(a[j]!=j-i)tmp++;
b[i][j]=tmp;
}
}
mem(f,inf);f[0][0]=0;
rep(i,1,n)
rep(j,1,i)
rep(k,0,i-1)f[i][j]=min(f[i][j],f[k][j-1]+b[k+1][i]);
rep(i,1,n)printf("%d\n",f[n][i]);
return 0;
}