UVALive 6855 Banks （暴力）

Banks

题目链接：

http://acm.hust.edu.cn/vjudge/contest/130303#problem/A

Description

http://7xjob4.com1.z0.glb.clouddn.com/8ce645bf3da25e2731b2fea4c21a985b

Input

The input file contains several test cases, each of them as described below.
On the first line, we have the number n of banks. On the second line, we have the capitals ki
(n > i ≥ 0) of all banks, in the order in which they are found on Wall Street from Wonderland. Each
capital is separated by a single whitespace from the next one, except for the final capital which is
directly followed by the newline character.

Output

For each test case, the output contains a single line with the value of the minimal number of magic
moves.

Sample Input

```
4
1 -2 -1 3
```

Sample Output

```
9
```

Source

2016-HUST-线下组队赛-4


##题意：

给出一个循环序列，每次可以操作可以把一个负数取反成a，并把其周围的两个数减去a.
求最少次数使得结果序列非负.


##题解：

如果序列能够达到全部非负的状态，那么无论先操作哪个数都是一样的次数.
所以直接暴力枚举所有负数，递归处理即可.


##代码：
``` cpp
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define LL long long
#define maxn 10100
#define inf 0x3f3f3f3f
#define mod 1000000007
#define mid(a,b) ((a+b)>>1)
#define eps 1e-8
#define IN freopen("in.txt","r",stdin);
using namespace std;

int num[maxn];

int ans, n;

void dfs(int cur) {

if(num[cur] >=0) return ;

num[cur] = -num[cur]; ans++;

int l = cur - 1; if(l == 0) l = n;

int r = cur + 1; if(r == n+1) r = 1;

num[l] -= num[cur];

num[r] -= num[cur];

if(num[l] < 0) dfs(l);

if(num[r] < 0) dfs(r);

}

int main()

{

//IN;

while(scanf("%d", &n) != EOF)
{
    for(int i=1; i<=n; i++) {
        scanf("%d", &num[i]);
    }

    ans = 0;
    for(int i=1; i<=n; i++) {
        if(num[i] < 0) {
            dfs(i);
        }
    }

    printf("%d\n", ans);
}

return 0;

}