Flexible application of POJ 3662 telephone Lines SPFA algorithm

Test instructions

to a graph with n nodes, which requires a path from 1 to n, you can make the K bar free, the cost of the path is the maximum of the length of the remaining edge of the path, and now requires the minimum value to be spent.

Ideas:

This problem can first think of the maximum value of the binary enumeration path, I think with SPFA more concise. The essence of SPFA is a search algorithm, and since it is a search, it involves the transfer of state. In the SPFA algorithm, which is generally the shortest path, when the node U is reached, the following transfer is required for E (U,V): if (D[v]>d[u]+w (e)) D[v]=d[u]+w (e). In the general case, to the junction U, to E (u,v) need to do a variety of transfer, such as the question to consider let e free and do not let e free two cases, specific please refer to the implementation.

Code:

POJ 3662//sepnine#include <iostream> #include <queue>using namespace std;const int Maxn=1024;const int Maxm=10024;struct edge{int V,w,next;} Edge[maxm*2];struct node{int u,used;}; int n,p,k,e;int head[maxn],dis[maxn][maxn],vis[maxn][maxn];void SPFA (int s,int t,int k) {queue<node> Q;memset ( Vis,0,sizeof (VIS)); int i,j;for (i=1;i<=n;++i) for (j=0;j<=k;++j) Dis[i][j]=int_max; Node x;x.u=s;x.used=0;dis[s][0]=0;vis[s][0]=1; Q.push (x); while (! Q.empty ()) {Node x=q.front (); int u=x.u,used=x.used; Q.pop (); vis[u][used]=0;for (int i=head[u];i!=-1;i=edge[i].next) {int v=edge[i].v,w=edge[i].w;if (used+1<=k& &dis[v][used+1]>dis[u][used]) {//Let this edge free, free number of sides used to +1dis[v][used+1]=dis[u][used];if (vis[v][used+1]==0) { Node x;x.u=v;x.used=used+1;vis[x.u][x.used]=1; Q.push (x);}} if (Dis[v][used]>max (dis[u][used],w)) {//Do not let this edge free, dis[v][used] is dis[u][used] and W is the maximum value Dis[v][used]=max (dis[u][used ],W); if (vis[v][used]==0) {Node x;x.u=v;x.used=used;vis[x.u][x.used]=1; Q.push (x);}}} return;} int main () {scanf("%d%d%d", &n,&p,&k); E=0;memset (Head,-1,sizeof (head)); while (p--) {int a,b,c;scanf ("%d%d%d", &a,&b,&c); edge[e].v=b;edge[e].w=c,edge[e].next=head[a];head[a]= e++;edge[e].v=a;edge[e].w=c;edge[e].next=head[b];head[b]=e++;} SPFA (1,n,k), int ans=int_max;for (int i=0;i<=k;++i) ans=min (Ans,dis[n][i]), if (Ans==int_max) printf ("-1");  elseprintf ("%d", ans); return 0;}

Flexible application of POJ 3662 telephone Lines SPFA algorithm