poj1797 Heavy Transportation SPFA Deformation __POJ

The following:

Given a non-direction graph, the maximum value of the minimum value from the path of 1 to n is obtained.

That is, there may be more than one path from 1 to n, and each path has an edge with the smallest weight, asking the maximum value of those edges.

idea:

The previously SPFA dist Array records the distance, and now only changes to the minimum value from the source point to I.

The original SPFA extension node:

for (int i = 0; i < edges[cur].size (); i++) {
	Edge e = edges[cur][i];
	if (<strong>dist[cur] + e.v < dist[e.to]</strong>) {
		dist[e.to] = dist[cur] + e.v
		if (!vis[e.to)) {
			vis[e.to] = 1;
			Q.push (e.to);}}

Now just change it to:

for (int i = 0; i < edges[cur].size (); i++) {
	Edge e = edges[cur][i];
	if (<strong>min (Dist[cur], e.v) > dist[e.to]</strong>) {
		dist[e.to] = min (dist[cur), e.v);
		if (!vis[e.to]) {
			vis[e.to] = 1;
			Q.push (e.to);}}

Can.

Code (1836K,282MS):

#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <

Vector> #include <queue> using namespace std;
	struct edge{int to, V;

Edge (int A, int b): to (a), V (b) {}};
int T;
int n, m; Vector<edge> edges[1005];
adjacency table int vis[1005];

int dist[1005];
	int SPFA () {queue<int> q;
	memset (Vis, 0, sizeof (VIS));
	memset (Dist, 0, sizeof (Dist));
	Q.push (1);
	VIS[1] = 1;
	DIST[1] = 0x3f3f3f3f;
		while (!q.empty ()) {int cur = q.front ();
		Q.pop ();
			for (int i = 0; i < edges[cur].size (); i++) {Edge e = edges[cur][i];
				if (min (dist[cur], e.v) > Dist[e.to]) {dist[e.to] = min (dist[cur), e.v);
					if (!vis[e.to]) {vis[e.to] = 1;
				Q.push (e.to);
	}} Vis[cur] = 0;
return dist[n];
	int main () {scanf ("%d", &t);
	int cas = 1;
		while (t--) {scanf ("%d%d", &n, &m);
		for (int i = 0; I <= N; i++) edges[i].clear ();
		int A, b, C; for (int i = 0; i < m; I+ +) {scanf ("%d%d%d", &a, &b, &c);
			Edges[a].push_back (Edge (b, c));
		Edges[b].push_back (Edge (A, c));
		int ans = SPFA ();
	printf ("Scenario #%d:\n%d\n\n", cas++, ans);
return 0;
 }