UVALive - 4255-Guess(拓扑排序)

Problem UVALive - 4255-Guess

Time Limit: 3000 mSec

UVALive - 4255-Guess(拓扑排序) Problem Description

UVALive - 4255-Guess(拓扑排序)

 

Input

The input consists of T test cases. The number of test cases T is given in the first line of the input. Each test case consists of two lines. The first line contains an integer n (1 ≤ n ≤ 10), where n is the length of a sequence of integers. The second line contains a string of n(n + 1)/2 characters such that the first n characters correspond to the first row of the sign matrix, the next n−1 characters to the second row, ..., and the last character to the n-th row.

UVALive - 4255-Guess(拓扑排序)Output

For each test case, output exactly one line containing a sequence of n integers which generates the sign matrix. If more than one sequence generates the sign matrix, you may output any one of them. Every integer in the sequence must be between −10 and 10, both inclusive.  

UVALive - 4255-Guess(拓扑排序)Sample Input

3 4 -+0++++--+ 2 +++ 5 ++0+-+-+--+-+-

UVALive - 4255-Guess(拓扑排序) Sample Output

-2 5 -3 1

3 4

1 2 -3 4 -5

 

题解:这个题联想到拓扑排序不容易(做题太少),不过有了这个大方向之后就很好办了,关于赋值的问题,给最开始入度为0的点随便赋个值,之后每一层的点的是上一层的值减一,最后把sum[0]要化为0, 因此每个数都减去sum[0]即可。

 

  1 #include <bits/stdc++.h>
  2 
  3 using namespace std;
  4 
  5 #define REP(i, n) for (int i = 1; i <= (n); i++)
  6 #define sqr(x) ((x) * (x))
  7 
  8 const int maxn = 50 + 10;
  9 const int maxm = 30 + 10;
 10 const int maxs = 10000 + 10;
 11 
 12 typedef long long LL;
 13 typedef pair<int, int> pii;
 14 typedef pair<double, double> pdd;
 15 
 16 const LL unit = 1LL;
 17 const int INF = 0x3f3f3f3f;
 18 const LL mod = 1000000007;
 19 const double eps = 1e-14;
 20 const double inf = 1e15;
 21 const double pi = acos(-1.0);
 22 
 23 int n, deg[maxn];
 24 vector<int> G[maxn];
 25 int ans[maxn];
 26 
 27 void toposort()
 28 {
 29     queue<int> que;
 30     for (int i = 0; i <= n; i++)
 31     {
 32         if (!deg[i])
 33         {
 34             que.push(i);
 35             ans[i] = 10;
 36         }
 37     }
 38 
 39     while (!que.empty())
 40     {
 41         int head = que.front();
 42         que.pop();
 43         for (auto v : G[head])
 44         {
 45             deg[v]--;
 46             if (!deg[v])
 47             {
 48                 que.push(v);
 49                 ans[v] = ans[head] - 1;
 50             }
 51         }
 52     }
 53 }
 54 
 55 int main()
 56 {
 57     ios::sync_with_stdio(false);
 58     cin.tie(0);
 59     freopen("input.txt", "r", stdin);
 60     //freopen("output.txt", "w", stdout);
 61     int T;
 62     cin >> T;
 63     while (T--)
 64     {
 65         cin >> n;
 66         for (int i = 0; i <= n; i++)
 67         {
 68             G[i].clear();
 69             deg[i] = ans[i] = 0;
 70         }
 71         string str;
 72         cin >> str;
 73         int pos = 0;
 74         for (int i = 1; i <= n; i++)
 75         {
 76             for (int j = i; j <= n; j++)
 77             {
 78                 if (str[pos] == '+')
 79                 {
 80                     G[j].push_back(i - 1);
 81                     deg[i - 1]++;
 82                 }
 83                 else if (str[pos] == '-')
 84                 {
 85                     G[i - 1].push_back(j);
 86                     deg[j]++;
 87                 }
 88                 pos++;
 89             }
 90         }
 91         toposort();
 92         for (int i = 1; i <= n; i++)
 93         {
 94             ans[i] -= ans[0];
 95         }
 96         ans[0] = 0;
 97         for (int i = 1; i <= n; i++)
 98         {
 99             cout << ans[i] - ans[i - 1];
100             if (i != n)
101             {
102                 cout << " ";
103             }
104             else
105             {
106                 cout << endl;
107             }
108         }
109     }
110     return 0;
111 }

 

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