[Usaco] wormholes [brute force]

Question: Give 2 k points on the plane, and pair them one by one. Ask how many pairing Methods There are to keep moving from a certain point to the right in the loop (k <= 6)

Idea: Because K is very small, the matching method is violent enumeration, and then ring judgment, when ring judgment, you need to pay attention to the four points 1, 2, 3, 1 and 3 on a straight line, pairing 2 and 4, you can find that it does not constitute a ring. For details, see the code.

 

/*{

ID: a4298442

Prob: Wormhole

Lang: C ++

}

*/

# Include <stdio. h>

# Include <iostream>

# Include <fstream>

# Include <string. h>

# Include <algorithm>

# Define maxn1000

Using namespace STD;

Ifstream fin ("wormhole. In ");

Ofstream fout ("wormhole. Out ");

Struct t {int X; int y;} p [maxn];

Int ans = 0, G [maxn], Match [maxn], n, visit [maxn];

Int CMP (t x, t y ){

Return (x. y <Y. Y | (x. y = Y. Y) & (x. x <Y. X )));

}

Int circle (int u)

{

Int vis [100] = {0}, V = G [u];

Vis [u] = 1;

While (V! = 0 ){

U = match [v];

If (! U) return 0; V = G [u];

If (vis [u]) return 1; vis [u] = 1;

}

Return 0;

}

Void DFS (int K, int U, int num ){

If (num = N ){

For (INT I = 1; I <= N; I ++) if (Circle (I) {ans ++; break ;}

Return;

}

Visit [k] = 1;

For (INT I = (u = 0 )? K + 1:1; I <= N; I ++) if (visit [I] = 0 ){

If (u = 0) {G [I] = K; G [k] = I ;}

DFS (I, u ^ 1, num + 1 );

If (u = 0) g [I] = G [k] = 0; else break;

}

Visit [k] = 0;

}

Int main (){

Fin> N;

For (INT I = 1; I <= N; I ++) Fin> P [I]. x> P [I]. Y;

Sort (p + 1, p + 1 + N, CMP );

For (INT I = 2; I <= N; I ++) if (P [I]. Y = P [I-1]. Y) Match [I-1] = I;

DFS (1, 0, 1 );

Fout <ans <Endl;

Return 0;

}

[Usaco] wormholes [brute force]