Problem C
FLYING TO FREDERICTON
After being inspired by Salvador Dali's artwork, Brett decided he would like to travel to Fredericton, New Brunswick to visit the Beaverbrook Art Gallery.
Brett lives in Calgary, and would like to find the cheapest flight or flights that would take him to Fredericton. He knows that a direct flight from Calgary to Fredericton, if one exists, would be absurdly expensive, so he is willing to tolerate a certain number of stopovers. However, there are multiple airlines in Canada with so many different flights between different cities now, which makes it very difficult for Brett to find the least expensive way to Fredericton! Can you write a program to help Brett plan his route?
Map showing a sample of the flights that would take Brett to Fredericton.
You will be given a list of cities between and including Calgary and Fredericton. The cities will be given in order of "nearest" to "farthest". The first city will always be "Calgary" and the last "Fredericton".
You will also be given a list of flights between pairs of cities, and the associated cost for each flight, taxes included. There will never be a flight from a farther city to a nearer city - Brett has already discarded those flights, deeming them to be a waste of time and money. Bear in mind, however, that there may be more than one flight between any two cities, as Brett is considering flights from all airlines.
Finally, you are presented with a number of queries. Each query is a single integer indicating the maximum number of stopovers Brett will tolerate. For each query, your program must calculate the least total cost of flights that would take Brett from Calgary to Fredericton with no more than the requested number of stopovers.
Input
The first line of input contains a single number indicating the number of scenarios to process. A blank line precedes each scenario.
Each scenario begins with a number N (2 ≤ N ≤ 100), the number of cities, followed by N lines containing the names of the cities. A city name is a string of up to 20 uppercase and lowercase letters. Next is a number M (0 ≤ M ≤ 1000), the number of flights available, followed by M lines describing the flights. Each flight is described by its departure city, its destination city, and an integer representing its cost in dollars. The final line starts with Q (1 ≤ Q ≤ 10), the number of queries, followed by Q numbers indicating the maximum number of stopovers.
Output
For each scenario, your program should output the scenario number, followed by the least total cost of the flights for each query. Follow the format of the sample output. If no flights can satisfy a query, write "No satisfactory flights". Output a blank line between scenarios.
Sample Input
2 4 Calgary Winnipeg Ottawa Fredericton 6 Calgary Winnipeg 125 Calgary Ottawa 300 Winnipeg Fredericton 325 Winnipeg Ottawa 100 Calgary Fredericton 875 Ottawa Fredericton 175 3 2 1 0 3 Calgary Montreal Fredericton 2 Calgary Montreal 300 Montreal Fredericton 325 1 0
Output for the Sample Input
Scenario #1 Total cost of flight(s) is $400 Total cost of flight(s) is $450 Total cost of flight(s) is $875 Scenario #2 No satisfactory flights---------------
占坑
---------------
#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
#include <algorithm>
#include <queue>
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=211;
const int maxm=2111;
map<string,int>mp;
char name[111111];
int S,T;
struct EdgeNode{
int to;
int w;
int next;
};
struct BellmanFord{
EdgeNode edges[maxm];
int head[maxn],edge,n;
bool inq[maxn][maxn];
queue< pair<int,int> >que;
int dis[maxn][maxn];
void addedge(int u,int v,int c){
edges[edge].w=c,edges[edge].to=v,edges[edge].next=head[u],head[u]=edge++;
}
void init(int n){
memset(head,-1,sizeof(head));
edge=0;
this->n=n;
}
bool spfa(int s,int len){
int u;
memset(dis,INF,sizeof(dis));
memset(inq,0,sizeof(inq));
while (!que.empty()) que.pop();
dis[0][s]=0;
inq[0][s]=true;
que.push( make_pair(0,s) );
while (!que.empty()){
int tim=que.front().first;
u=que.front().second;
que.pop();
inq[tim][u]=false;
for (int i=head[u];i!=-1;i=edges[i].next){
int v=edges[i].to;
int w=edges[i].w;
if ( tim+1<=len&&dis[tim+1][v]>dis[tim][u]+w ){
dis[tim+1][v]=dis[tim][u]+w;
//do something
//nice try
if (!inq[tim+1][v]){
inq[tim+1][v]=true;
que.push( make_pair(tim+1,v) );
}
}
}
}
return true;
}
}solver;
int n,m;
int main()
{
int TMD;
int cas=0;
scanf("%d",&TMD);
while (TMD--){
//input()
scanf("%d",&n);
solver.init(n);
for (int i=1;i<=n;i++){
scanf("%s",name);
mp[name]=i;
}
scanf("%d",&m);
for (int i=0;i<m;i++){
int x,y,c;
scanf("%s",name);
x=mp[name];
scanf("%s",name);
y=mp[name];
scanf("%d",&c);
solver.addedge(x,y,c);
}
S=mp["Calgary"];
T=mp["Fredericton"];
//Solve()
int Q,qst;
printf("Scenario #%d
",++cas);
scanf("%d",&qst);
while (qst--){
scanf("%d",&Q);
Q++;
if (Q>n) Q=n;
if (Q<1) Q=1;
solver.spfa(S,Q);
int ans=INF;
for(int i=0;i<=Q;i++)
ans=min(ans,solver.dis[i][T]);
if (ans==INF){
printf("No satisfactory flights
");
}else{
printf("Total cost of flight(s) is $%d
",ans);
}
}
if (TMD!=0) puts("");
}
return 0;
}