两次SPFA
第一关找:从1没有出发点到另一个点的多少是留给油箱
把边反过来再找一遍:重每一个点到终点最少须要多少油
Greedy Driver
Time Limit: 2 Seconds Memory Limit: 65536 KB
Edward is a truck driver of a big company. His daily work is driving a truck from one city to another. Recently he got a long distance driving work, so he may be needed to add fuel to
his truck on the way. His boss gives him a fuel card, he can use the card anytime at any fuel station and add any amount of fuel. He does not spend any money when he uses the card. He noticed that there are some cities where he could sell fuel. The greedy
driver Edward want to make some money while he doing this work. And this is his plan:
There are N cities on his map and numbered from 1 to N, he need to drive from city 1 to city N to finish this task. Driving from one city to another need
fuel cost too. There are some of the cities has fuel station, he could use his fuel card at these cities. Notice his truck has a fuel tank capacity C and it could not exceed the capacity when add fuel. At the beginning, he is at city 1 and
he has a full tank.
At some cities he could sell any amount of fuel he has, and the prices at each city maybe different. Since he does not want to be found, he sell the fuel at most once totally. Under the
premise of driving to city N finally, he wants to make the maximal money.
Input
There are multiple test cases. For each test case:
The first line contains three integer N (1 ≤ N ≤ 1000), M (1 ≤ M ≤ 100000), C(1 ≤ C ≤ 30000), N is the number
of cities, M is the number of roads, please notice that the roads is single-way, and C is the fuel tank capacity.
The after M lines describe the roads, each line has three integers A, B (1 ≤ A, B ≤ N), L (1 ≤ L ≤
30000). That means the fuel cost from city A to city B is L.
Then a single line contain one integer P (0 ≤ P ≤ N), it is the number of cities which has a fuel station.
The after single line contains P integers, each integer pi (1 ≤ pi ≤ N) is a city number means this city has
a fuel station.
Then a single line contain one integer Q (0 ≤ Q ≤ N), it is the number of cities which he could sell fuel.
Each of the after Q lines contains two integers qi (1 ≤ qi ≤ N), vi (1 ≤ vi ≤
30000), means the price at city qi is vi. If he sell one unit of fuel, he will get vi money.
There is a blank line between every two cases.
Process to the end of input.
Output
One line for each test case. The maximal money he could make while doing this task, or -1 if he could not reach city N.
Sample Input
5 6 10
1 2 4
1 4 1
4 3 1
2 5 1
4 5 2
3 2 1
1
3
1
2 2
Sample Output
16
Author: LI, Huang
Source: ZOJ Monthly, June 2014
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue> using namespace std; const int maxn=110000;
const int INF=0x3f3f3f3f; struct Edge
{
int to,next,cost;
}edge[maxn],edge2[maxn]; int Adj[maxn/100],Adj2[maxn/100],Size,Size2;
int fuel[maxn/100],price[maxn/100],n,m,c,p,q; void init()
{
Size=Size2=0;
memset(Adj,-1,sizeof(Adj));
memset(Adj2,-1,sizeof(Adj2));
memset(fuel,0,sizeof(fuel));
memset(price,0,sizeof(price));
} void Add_Edge(int u,int v,int w)
{
edge[Size].next=Adj[u];
edge[Size].to=v;
edge[Size].cost=w;
Adj[u]=Size++;
} void Add_Edge2(int u,int v,int w)
{
edge2[Size2].next=Adj2[u];
edge2[Size2].to=v;
edge2[Size2].cost=w;
Adj2[u]=Size2++;
} int dmax[maxn/100],cq[maxn],dmin[maxn/100];
bool inq[maxn/100]; bool spfa_find_max()
{
memset(dmax,-1,sizeof(dmax));
memset(cq,0,sizeof(cq));
memset(inq,false,sizeof(inq));
dmax[1]=c;
queue<int> q;
inq[1]=true,cq[1]=1; q.push(1); while(!q.empty())
{
int u=q.front();q.pop();
for(int i=Adj[u];~i;i=edge[i].next)
{
int v=edge[i].to;
if(dmax[u]-edge[i].cost<0) continue;
if(dmax[v]<dmax[u]-edge[i].cost)
{
if(fuel[v]) dmax[v]=c;
else dmax[v]=dmax[u]-edge[i].cost;
if(inq[v]==false)
{
inq[v]=true;
cq[v]++;
if(cq[v]>=n) return false;
q.push(v);
}
}
}
inq[u]=false;
}
return true;
} bool spfa_find_min()
{
memset(dmin,63,sizeof(dmin));
memset(inq,false,sizeof(inq));
memset(cq,0,sizeof(cq));
dmin[n]=0;
queue<int> q;
q.push(n);
inq[n]=true,cq[n]=1; while(!q.empty())
{
int u=q.front(); q.pop(); for(int i=Adj2[u];~i;i=edge2[i].next)
{
int v=edge2[i].to;
if(dmin[u]+edge2[i].cost>c) continue;
if(dmin[v]>dmin[u]+edge2[i].cost)
{
if(fuel[v]) dmin[v]=0;
else
{
dmin[v]=dmin[u]+edge2[i].cost;
}
if(inq[v]==false)
{
inq[v]=true;
cq[v]++;
if(cq[v]>=n) return false;
q.push(v);
}
}
}
inq[u]=false;
}
return true;
} int main()
{
while(scanf("%d%d%d",&n,&m,&c)!=EOF)
{
init();
for(int i=0;i<m;i++)
{
int u,v,x;
scanf("%d%d%d",&u,&v,&x);
Add_Edge(u,v,x);
Add_Edge2(v,u,x);
}
scanf("%d",&p);
for(int i=0;i<p;i++)
{
int x; scanf("%d",&x);
fuel[x]=1;
}
scanf("%d",&q);
for(int i=0;i<q;i++)
{
int x,y; scanf("%d%d",&x,&y);
price[x]=y;
}
spfa_find_max();///油箱里最多剩下多少油
spfa_find_min();///到终点最少须要多少油
if(dmax[n]==-1||dmin[1]==INF) puts("-1");
else
{
int ans=0;
for(int i=1;i<=n;i++)
{
if(price[i]&&dmax[i]!=-1&&dmin[i]!=INF)
{
ans=max(ans,price[i]*(dmax[i]-dmin[i]));
}
}
printf("%d\n",ans);
}
}
return 0;
}
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