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.

输入输出样例

6
-1 1 0 -1 0 1

3

说明

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

很板的一眼DP题，直接dp[i][3]表示在第i个位置的时候最后是0,-1,1的最小花费是多少就好了，总复杂度O(n)

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 1e6 + 5, inf = 0x3f3f3f3f;
int n, dp[maxn][3], num[maxn], ans;

int main() {
    scanf("%d", &n);
    memset(dp, 0x3f, sizeof(dp));
    for(register int i = 1; i <= n; ++i) scanf("%d", &num[i]);
    dp[1][num[1] + 1] = 0;
    for(register int i = 2; i <= n; ++i) {
        for(register int j = 0; j < 3; ++j) {
            if((j - 1) + num[i] < 2 && (j - 1) + num[i] > -2 && (j - 1) + num[i] >= j - 1)
                dp[i][j + num[i]] = min(dp[i][j + num[i]], dp[i - 1][j] + 1);
            if(2 * (j - 1) + num[i] < 2 && 2 * (j - 1) + num[i] > -2 && 2 * (j - 1) + num[i] >= j - 1)
                dp[i][2 * (j - 1) + num[i] + 1] = min(dp[i][2 * (j - 1) + num[i] + 1], dp[i - 1][j] + 2);
            if(num[i] >= j - 1) 
                dp[i][num[i] + 1] = min(dp[i][num[i] + 1], dp[i - 1][j]);
        }
    }
    ans = min(dp[n][0], min(dp[n][1], dp[n][2]));
    printf(ans == inf ? "BRAK" : "%d", ans);
    return 0;
}