题目描述
A sequence of 
integers 
from the set 
is given.
The bytecomputer is a device that allows the following operation on the sequence:
incrementing 
by 
for any 
.
There is no limit on the range of integers the bytecomputer can store, i.e., each 
can (in principle) have arbitrarily small or large value.
Program the bytecomputer so that it transforms the input sequence into a non-decreasing sequence (i.e., such that 
) with the minimum number of operations.
给一个只包含-1,0,1的数列，每次操作可以让a[i]+=a[i-1]，求最少操作次数使得序列单调不降
输入输出格式
输入格式：
The first line of the standard input holds a single integer 
(
), the number of elements in the (bytecomputer‘s) input sequence.
The second line contains 
integers 
(
) that are the successive elements of the (bytecomputer‘s) input sequence, separated by single spaces.
In tests worth 24% of the total points it holds that 
, and in tests worth 48% of the total points it holds that 
.
输出格式：
The first and only line of the standard output should give one integer, the minimum number of operations the bytecomputer has to perform to make its input sequence non-decreasing, of the single word BRAK (Polish for none) if obtaining such a sequence is impossible.
Solution
DP。
比较明显的是我们最多也只需要把一个数位加到1，最少减到-1就可以了。用$f[i][j],i\in Z,1\le i \le n,j\in Z,0 \le j \le 2 $表示第i个数的第j状态需要怎么转移过来，然后暴力枚举之前的可能情况，然后直接根据当前的情况进行转移就可以了。
Code
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <cstdlib>
#include <cmath>
#include <string>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
#define re register
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define ms(arr) memset(arr, 0, sizeof(arr))
const int inf = 0x3f3f3f3f;
int f[1000001][3],n,m,a[1000001],ans;
inline int read()
{
int x=0,c=1;
char ch=' ';
while((ch>'9'||ch<'0')&&ch!='-')ch=getchar();
while(ch=='-') c*=-1,ch=getchar();
while(ch<='9'&&ch>='0')x=x*10+ch-'0',ch=getchar();
return x*c;
}
int main()
{
//freopen("date.in","r",stdin);
n=read();
memset(f,126,sizeof(f));
for(re int i=1;i<=n;i++)
a[i]=read();
f[1][a[1]+1]=0;
for(re int i=2;i<=n;i++){
if(a[i]==-1){
f[i][0]=f[i-1][0];
f[i][2]=f[i-1][2]+2;
}else if(a[i]==0){
f[i][0]=f[i-1][0]+1;
f[i][1]=min(f[i-1][0],f[i-1][1]);
f[i][2]=f[i-1][2]+1;
}else{
f[i][0]=f[i-1][0]+2;
f[i][1]=f[i-1][0]+1;
f[i][2]=min(min(f[i-1][0],f[i-1][1]),f[i-1][2]);
}
}
ans=min(min(f[n][0],f[n][1]),f[n][2]);
if(ans>200000000) cout<<"BRAK";
else cout<<ans;
return 0;
}原文地址：https://www.cnblogs.com/victorique/p/8877070.html