UVALive 6855 Banks （暴力）

Banks

题目连接：

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，每次能够操做能够把一个负数取反成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：

若是序列可以达到所有非负的状态，那么不管先操做哪一个数都是同样的次数.
因此直接暴力枚举全部负数，递归处理便可.

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#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;
}