Shipping Routes

FJNU.1689

Description
The Slow Boat to China Shipping company needs a program to help them quickly quote costs to prospective customers. The cost of a shipment depends on the size of the shipment and on how many shipping legs it requires. A shipping leg connects two warehouses, but since every pair of warehouses is not directly connected by a leg, it might require more than one leg to send a shipment from one warehouse to another.
A data set can represent from 1 to 30 warehouses. A two-letter code name will identify each warehouse (capital letters only). Shipping legs can exist between any two distinct warehouses. All legs are bidirectional.
The cost of a shipment is equal to the size of the shipment times the number of shipping legs required times $100.
The input to the program identifies the warehouse code names and the existence of all shipping legs. For a given shipping request, consisting of the size of the shipment, the source warehouse and the destination warehouse, the program will output the best (cheapest) cost for the shipment, if it is possible to send shipments from the requested source to the requested destination. Alternately, the program must state that the request cannot be fulfilled.

Input
The first line will contain an integer from 1 to 10 inclusive that represents the number of data sets in the input file. Each data set represents a new shipping configuration.
The first line of data in a data set will contain three integers, say M, N, and P: M is an integer from 1 to 30 inclusive representing the number of warehouses in the data set; N is an integer from 0 to M*(M-1)/2 inclusive that represents the number of legs between warehouses in the data set; P is an integer from 0 to 10 inclusive that represents the number of shipping requests for which cost information is required.
The second line of data in a data set contains M two-letter code names for the M warehouses of the data sets. Only capital letters are used. A single blank separates code names.
N lines follow the line of code names, containing shipping leg information in the format: ``XX YY", with XX and YY being the codes for two distinct warehouses in the set that have a direct link (a shipping leg) between them. There will be a single blank between the warehouse codes.
The N lines of shipping leg information are followed by P lines of shipping requests, one request per line. Each shipping request will begin with an integer between 1 and 20 inclusive that represents the size of the shipment. The shipment size will be followed by a pair of code names in the format ``AA BB", with AA and BB being the code for two distinct warehouses in the set which represent the source and destination of the requested shipment.
The input will be valid and consistent. A shipping leg will only be represented once within a data set. Data about legs will always refer to warehouses that have been identified as belonging to the data set. See the example below for clarification of the input format.

Output
The first line of output should read ``SHIPPING ROUTES OUTPUT". For each data set there should be a section of output enumerating which data set the output section represents followed by P lines of the required information. Each of the P lines should list either the cheapest cost of the respective shipment or the phrase ``NO SHIPMENT POSSIBLE". The end of the output should be noted also. Produce output consistent with the example below.

Sample Input
2
6 7 5
AA CC QR FF DD AB
AA CC
CC QR
DD CC
AA DD
AA AB
DD QR
AB DD
5 AA AB
14 DD CC
1 CC DD
2 AA FF
13 AB QR
3 0 1
AA BB CC
5 AA CC

Sample Output
SHIPPING ROUTES OUTPUT

DATA SET 1

$500
$1400
$100
NO SHIPMENT POSSIBLE
$2600

DATA SET 2

NO SHIPMENT POSSIBLE

END OF OUTPUT

Source
Mid-Atlantic USA 1996

My Program

#include<iostream>
#include<map>
#include<string>
#define N 31
using namespace std;
bool data[N][N];
int n,visit[N];

int shortest(int s,int e)
...{
    int front,rear,queue[N],t,i;
    memset(visit,0,sizeof(visit));
    queue[0]=s;front=0;rear=1;visit[s]=0;
    while(front<rear)
    ...{
        t=queue[front];
        for(i=0;i<n;i++)
            if(data[t][i]&&visit[i]==0&&i!=s)
            ...{
                visit[i]=visit[t]+1;
                queue[rear++]=i;
            }
        if(data[t][e])return visit[e];
        front++;
    }
    return 0;
}

int main()
...{
    string name,to;
    map<string,int>id;
    int k,m,p,t,i,j,s;
    cin>>k;
    cout<<"SHIPPING ROUTES OUTPUT"<<endl<<endl;
    for(t=1;t<=k;t++)
    ...{
        cout<<"DATA SET "<<t<<endl<<endl;
        cin>>n>>m>>p;
        memset(data,false,sizeof(data));
        for(i=0;i<n;i++)
        ...{
            cin>>name;
            id[name]=i;
        }
        while(m--)
        ...{
            cin>>name>>to;
            i=id[name];j=id[to];
            data[i][j]=data[j][i]=true;
        }
        while(p--)
        ...{
            cin>>m>>name>>to;
            i=id[name];j=id[to];
            s=shortest(i,j);
            if(s==0)
                cout<<"NO SHIPMENT POSSIBLE"<<endl;
            else
                cout<<"$"<<s*m*100<<endl;
        }
        cout<<endl;
    }
    cout<<"END OF OUTPUT"<<endl;
    return 0;
}

 

YOYO's Note:
┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄┄它是华丽的分隔线

【题意简述】

在n个码头（最多30个）间有m条水路，
我们要走p次，每次走从某一码头到另一码头k次，
不论走哪条，每个线段都是$100，
求最少消耗。

【粗略分析】

图论。求其中一点到另一点的最短路径，
不过这里它用char们来表示码头，因此加个map用来随时读存码头编号。
接着简单BFS搜索路径，未有则no shipment possible，
有则输出长度*100即为最短消耗。

【C++源代码】

#include<iostream>
#include<map>
#include<string>
#define N 31
using namespace std;
bool data[N][N];
int n,visit[N];

int shortest(int s,int e)
...{
    int front,rear,queue[N],t,i;
    memset(visit,0,sizeof(visit));
    queue[0]=s;front=0;rear=1;visit[s]=0;
    while(front<rear)
    ...{
        t=queue[front];
        for(i=0;i<n;i++)
            if(data[t][i]&&visit[i]==0&&i!=s)
            ...{
                visit[i]=visit[t]+1;
                queue[rear++]=i;
            }
        if(data[t][e])return visit[e];
        front++;
    }
    return 0;
}

int main()
...{
    string name,to;
    map<string,int>id;
    int k,m,p,t,i,j,s;
    cin>>k;
    cout<<"SHIPPING ROUTES OUTPUT"<<endl<<endl;
    for(t=1;t<=k;t++)
    ...{
        cout<<"DATA SET "<<t<<endl<<endl;
        cin>>n>>m>>p;
        memset(data,false,sizeof(data));
        for(i=0;i<n;i++)
        ...{
            cin>>name;
            id[name]=i;
        }
        while(m--)
        ...{
            cin>>name>>to;
            i=id[name];j=id[to];
            data[i][j]=data[j][i]=true;
        }
        while(p--)
        ...{
            cin>>m>>name>>to;
            i=id[name];j=id[to];
            s=shortest(i,j);
            if(s==0)
                cout<<"NO SHIPMENT POSSIBLE"<<endl;
            else
                cout<<"$"<<s*m*100<<endl;
        }
        cout<<endl;
    }
    cout<<"END OF OUTPUT"<<endl;
    return 0;
}

【注意事项】

※ 输出格式

【点评】

很意外地居然一次过 = =