USACO4.1.3 Fence Loop Non-direction graph minimum ring __usaco

The idea is to ask you to ask for the smallest loop of the graph, but the way to get the data is disgusting.

I use and check the collection + violence of the way ...

First the vertices of each edge are labeled, and then ... The point where a can go to b,b can also go to the side of a, giving and for a point ...

Then Floyd to find the smallest ring.

Floyd the smallest ring I don't quite understand. But use it first and think again in class.

   executing ... Test 1:test OK [0.005 secs, 4028 KB]
   test 2:test OK [0.003 secs, 4028 KB]
   test 3:test OK [0.005 secs, 4028 kb]< C3/>test 4:test OK [0.008 secs, 4028 KB]
   test 5:test OK [0.005 secs, 4028 KB]
   test 6:test OK [0.008 secs, 4028 KB]
   Test 7:test OK [0.014 secs, 4028 KB]
   test 8:test OK [0.049 secs, 4028 KB]
   test 9:test OK [0.041 secs, 4 028 KB] All

tests OK.

/* Task:fence6 lang:c++ * * #include <iostream> #include <cstring> #include <cstdio> #include <cmath&
Gt

#include <vector> using namespace std;
int n;
	struct Edge {int Length;
	int q1_out, q2_out;
	vector<int>q1;
	vector<int>q2;

Edge () {} edge (int _length): Length (_length) {}}x[1000];
int father[1000]= {0};

int g[250][250], dist[250][250];
	int father_is_who (int k) {if (Father[k]) return father[k] = father_is_who (father[k));
else return k;
	} void Init () {scanf ("%d", &n);		
		for (int i = 1; I <= n; + + i) {int Num, Length, OUT1, Out2;
		scanf ("%d%d%d%d", &num, &length, &AMP;OUT1, &out2);
		X[num] = Edge (Length);
			for (int j = 0; J!= Out1; + + j) {int tmp;
			scanf ("%d", &tmp);
		X[num].q1.push_back (TMP);
			for (int j = 0; J!= Out2; + + j) {int tmp;	
			scanf ("%d", &tmp);
		X[num].q2.push_back (TMP); for (int i = 1; I <= n; + + i) {x[i].
		Q1_out = i * 2-1; X[i].
Q2_out = i * 2;	for (int i = 1; I <= n; + i) {for (int j = 0; J!= x[i].q1.size (); + + j) {//i from out1 Point, can go to x[i].q1[j]
			int would = X[i].q1[j];
				for (int k = 0; k!= x[will].q1.size (); + + K)//will departs from OUT1 to will_will {int will_will = x[will].q1[k]; if (Will_will = i) {int a = father_is_who (X[i].	
					Q1_out); int b = father_is_who (X[will].
					Q1_out);
				if (a!= b) father[a] = b;
				for (int k = 0; k!= x[will].q2.size (); + + K)//will departs from OUT1 to will_will {int will_will = x[will].q2[k]; if (Will_will = i) {int a = father_is_who (X[i].	
					Q1_out); int b = father_is_who (X[will].
					Q2_out);
				if (a!= b) father[a] = b;
			for (int j = 0; J!= x[i].q2.size (); + + j) {//i from Out2 point to x[i].q1[j] int will = x[i].q2[j];
				for (int k = 0; k!= x[will].q1.size (); + + K)//will departs from OUT1 to will_will {int will_will = x[will].q1[k]; if (Will_will = i) {int a = father_is_who (X[i].	
					Q2_out); Int b = father_is_who (X[will].
					Q1_out);
				if (a!= b) father[a] = b;
				for (int k = 0; k!= x[will].q2.size (); + + K)//will departs from OUT1 to will_will {int will_will = x[will].q2[k]; if (Will_will = i) {int a = father_is_who (X[i].	
					Q2_out); int b = father_is_who (X[will].
					Q2_out);
				if (a!= b) father[a] = b;
	}} memset (g, sizeof (g)); for (int i = 1; I <= n; + + i) {int a = father_is_who (X[i].
		Q1_out); int b = father_is_who (X[i].	
		Q2_out); G[A][B] = G[b][a] = x[i].
	Length; for (int i = 1; I <= 2 * n; + + i) for (int j = 1; J <= 2 * N; + + j) Dist[i][j] = g[i][j];//cout << g[i][	J] << "";

cout<<endl;
	} void doit () {int ans = 0x7fffffff; for (int k = 1; k <= 2 * n; + + K) {for (int i = 1; I!= K; + i) for (int j = i + 1; j!= K; + + j) if (dist[
		I][J] + g[j][k] + G[k][i] < ans ans = dist[i][j] + g[j][k] + g[k][i]; for (int i = 1; I <= 2 * n; + + i) for (int j = 1; j; = 2 * n;
	+ + j) if (Dist[i][k] + dist[k][j] < Dist[i][j]) dist[i][j] = Dist[i][k] + dist[k][j];
printf ("%d\n", ans);
	int main () {freopen ("fence6.in", "R", stdin);
	Freopen ("Fence6.out", "w", stdout);
	Init ();
	Doit ();
return 0;
 }