Application of POJ 3635 full Tank dynamic programming idea in SPFA algorithm

Test instructions

There are n cities, and M of the road known length, walking 1 units away from the road to spend 1 units of oil, each city can refuel and have their own oil price, now to two cities S,t, ask from S to t at least how much oil to spend.

Ideas:

Most of the online is to take priority queue to do, take SPFA do more interesting, but the general SPFA will time out. DIS[I][J] represents the minimum cost of J unit oil in the tank when it comes to city I. For city X, Y, their distance is W, if there is dis[x][j]<dis[y][i-w], then Y is determined to update, note if for each dis[x][j]<dis[y][j-w],dis[x][j+1]<dis[y][ J-w+1] ... All with the new dis[y][m],dis[y][m+1]...dis[y][c] words will time out, need to use the idea of dynamic planning in the determination of Y update only the total update once, specifically see the Code comment section.

Code:

#include <iostream> #include <queue>using namespace std;const int maxn=1024;const int Maxm=10024;const int Maxc=128;struct edge{int V,w,next;} Edge[maxm*2];int price[maxn];int head[maxn];int inq[maxn];int dis[maxn][maxc];int n,m,e;void spfa (int c,int s,int t) { int i,j;queue<int> Q;memset (inq,0,sizeof (INQ)); for (i=0;i<n;++i) for (j=0;j<=c;++j) Dis[i][j]=int_max; for (J=0;J&LT;=C;++J) dis[s][j]=price[s]*j;inq[s]=1; Q.push (s); while (! Q.empty ()) {int U=q.front (); Q.pop (); inq[u]=0;for (int i=head[u];i!=-1;i=edge[i].next) {int v=edge[i].v,w=edge[i].w;int flag=0;for (int j=w;j<= C;++J) if (Dis[u][j]<dis[v][j-w]) {dis[v][j-w]=dis[u][j]; flag=1;} if (flag==1) {for (int j=1;j<=c;++j)//With the idea of dynamic planning is updated only once Dis[v][j]=min (Dis[v][j],dis[v][j-1]+price[v]); if (inq[v]==0) { Inq[v]=1; Q.push (v);}} }/* Timeout Code: for (int i=head[u];i!=-1;i=edge[i].next) {int v=edge[i].v,w=edge[i].w;int flag=0;for (int j=w;j<=c;++j) if ( Dis[u][j]<dis[v][j-w]) {dis[v][j-w]=dis[u][j];for (int k=j-w;k<=c;++k)//try to update to full dis[v][k every time]=min (dis[v][k],dis[v][j-w]+ (k (j-w)) *price[v]); flag=1;} if (flag==1&&inq[v]==0) {inq[v]=1; Q.push (v);}} */}int Ans=int_max;for (i=0;i<=c;++i) ans=min (Ans,dis[t][i]), if (Ans==int_max) printf ("impossible\n"); elseprintf ("%d\n", ans); return;} int main () {E=0;memset (head,-1,sizeof (head)), scanf ("%d%d", &n,&m), int i;for (i=0;i<n;++i) scanf ("%d", &price[i]), while (m--) {int u,v,w;scanf ("%d%d%d", &u,&v,&w); edge[e].v=v;edge[e].w=w;edge[e].next= head[u];head[u]=e++;edge[e].v=u;edge[e].w=w;edge[e].next=head[v];head[v]=e++;} int q,c,s,t;scanf ("%d", &q), while (q--) {scanf ("%d%d%d", &c,&s,&t); SPFA (c,s,t);} return 0;}

Application of POJ 3635 full Tank dynamic programming idea in SPFA algorithm