Ultraviolet a La 4255 Guess (topological arrangement)

Ultraviolet a La 4255 Guess (topological arrangement)

Topology arrangement applies to Directed Acyclic graphs (dags.

Construct One-Way edges between all nodes.

In a specific problem, abstract the outbound vertex and edge (unidirectional edge). The unidirectional edge corresponds to the size relationship or demand relationship between specific points.

After creating a plot, all the relationships in the problem can be expressed by directed edges between points.

In this question.

(1) It is not easy to represent each number when it is constructed into a point.

The prefix and construction are made into points, and all the intervals and values can have two prefixes. The size relationship between different prefixes can be obtained from all intervals and positive and negative values or 0, that is, the unidirectional edge between points.

(2) Notice that it may be zero, indicating that the two prefix values are equal, that is, there may be two points in the graph into the ring.

Solution:

Dis [I] [j] = 0, indicating equal, 1 indicating I --> j,-1 representing j -- I.

When a vertex is obtained, all dis [] [] = 0 is taken and assigned the same value.

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             #include using namespace std;//LOOP#define FE(i, a, b) for(int i = (a); i <= (b); ++i)#define FD(i, b, a) for(int i = (b); i>= (a); --i)#define REP(i, N) for(int i = 0; i < (N); ++i)#define CLR(A,value) memset(A,value,sizeof(A))#define CPY(a, b) memcpy(a, b, sizeof(a))#define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)//INPUT#define RI(n) scanf("%d", &n)#define RII(n, m) scanf("%d%d", &n, &m)#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)#define RS(s) scanf("%s", s)//OUTPUT#define WI(n) printf("%d\n", n)#define WS(s) printf("%s\n", s)typedef long long LL;const int INF = 1000000007;const double eps = 1e-10;const int maxn = 20;int T, n;int suf[maxn];int dis[maxn][maxn];int du[maxn];vector
             
              a;void solve(){ int tot = 0; int val = 0; while (tot < n + 1) { for (int i = 0; i <= n; i++) { if (du[i] == 0) { a.clear(); a.push_back(i); suf[i] = val; for (int j = 0; j <= n; j++) { if (i != j && du[j] != -1 && dis[i][j] == 0) { suf[j] = val; a.push_back(j); } } val++; tot += a.size(); for (int j = 0; j < a.size(); j++) { int x = a[j]; for (int r = 0; r <= n; r++) if (du[r] != -1 && dis[x][r] == 1) du[r]--; } for (int j = 0; j < a.size(); j++) du[a[j]] = -1; } } } int x = suf[0]; for (int i = 0; i <= n; i++) suf[i] -= x; for (int i = 1; i <= n; i++) { if (i != 1) printf(" "); printf("%d", suf[i] - suf[i -1]); } puts("");}int main (){ scanf("%d",&T); while (T--) { scanf("%d", &n); memset(dis, 0, sizeof(dis)); memset(du, 0, sizeof(du)); for (int i = 1; i <= n; i++) { for (int j = i; j <= n; j++) { char x; scanf(" %c", &x); if (x == '-') { du[i - 1]++; dis[j][i - 1] = 1; dis[i - 1][j] = -1; } else if (x == '+') { du[j]++; dis[i - 1][j] = 1; dis[j][i - 1] = -1; } } } solve(); } return 0;}