HDU - 2923 - Einbahnstrasse

题目：

Einbahnstrasse

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1548    Accepted Submission(s): 428

Problem Description

Einbahnstra
e (German for a one-way street) is a street on which vehicles should
only move in one direction. One reason for having one-way streets is to
facilitate a smoother flow of traffic through crowded areas. This is
useful in city centers, especially old cities like Cairo and Damascus.
Careful planning guarantees that you can get to any location starting
from any point. Nevertheless, drivers must carefully plan their route in
order to avoid prolonging their trip due to one-way streets.
Experienced drivers know that there are multiple paths to travel between
any two locations. Not only that, there might be multiple roads between
the same two locations. Knowing the shortest way between any two
locations is a must! This is even more important when driving vehicles
that are hard to maneuver (garbage trucks, towing trucks, etc.)

You
just started a new job at a car-towing company. The company has a
number of towing trucks parked at the company's garage. A tow-truck
lifts the front or back wheels of a broken car in order to pull it
straight back to the company's garage. You receive calls from various
parts of the city about broken cars that need to be towed. The cars have
to be towed in the same order as you receive the calls. Your job is to
advise the tow-truck drivers regarding the shortest way in order to
collect all broken cars back in to the company's garage. At the end of
the day, you have to report to the management the total distance
traveled by the trucks.

 

Input

Your
program will be tested on one or more test cases. The first line of
each test case specifies three numbers (N , C , and R ) separated by one
or more spaces. The city has N locations with distinct names,
including the company's garage. C is the number of broken cars. R is
the number of roads in the city. Note that 0 < N < 100 , 0<=C
< 1000 , and R < 10000 . The second line is made of C + 1 words,
the first being the location of the company's garage, and the rest being
the locations of the broken cars. A location is a word made of 10
letters or less. Letter case is significant. After the second line,
there will be exactly R lines, each describing a road. A road is
described using one of these three formats:

A -v -> B
A <-v - B
A <-v -> B

A
and B are names of two different locations, while v is a positive
integer (not exceeding 1000) denoting the length of the road. The first
format specifies a one-way street from location A to B , the second
specifies a one-way street from B to A , while the last specifies a
two-way street between them. A , ``the arrow", and B are separated by
one or more spaces. The end of the test cases is specified with a line
having three zeros (for N , C , and R .)

The test case in the example below is the same as the one in the figure.

 

Output

For each test case, print the total distance traveled using the following format:

k . V

Where k is test case number (starting at 1,) is a space, and V is the result.

 

Sample Input

4 2 5

NewTroy Midvale Metrodale

NewTroy <-20-> Midval

Midvale --50-> Bakerline

NewTroy <-5-- Bakerline

Metrodale <-30-> NewTroy

Metrodale --5-> Bakerline

0 0 0

 

Sample Output

1. 80

 

　　题意：给你衣服一幅图，上面有一些点，一些有向边，一些无向边，以及边权，给你一个起点，一些目的地，目的地上面有一些破车，每一次你只可以到一个目的地回收一辆破车，到达目的地以后要把破车送回起点，求最终走的路程是多少。

　　代码敲出来以后各种卡，调试了好几遍以后终于把有的bug都去掉了，用c++交上去结果ce了，检查了很久发现原来少了个string的头文件，可是g++可以编译= =。加上去以后还是一直wa，后来瞄了一眼人家的题解，想起每一个目的地不止一辆破车，虽然这个一开始也想到，但是忽略了一次一辆拖车只可以拖一辆破车。然后把每个目的地的破车的数目乘上去以后就过了= =。

 

代码：

 

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <string>
 #include <map>
 #define INF 1000000000
 #define MAX 110
 using namespace std;

 map<string,int> M;

 int dis[MAX][MAX],amount[MAX],n;

 void Flyod()
 {
     int i,j,u;
     for(i=;i<=n;i++)
     {
         for(j=;j<=n;j++)
         {
             if(dis[i][j]<=) dis[i][j]=INF;
             if(i==j) dis[i][j]=;
         }
     }
     for(u=;u<=n;u++)
     {
         for(i=;i<=n;i++)
         {
             for(j=;j<=n;j++)
             {
                 int len=dis[i][u]+dis[u][j];
                 if(dis[i][j]>len) dis[i][j]=len;
             }
         }
     }
     /*
     for(i=1;i<=n;i++)
     {
         for(j=1;j<=n;j++)
         {
             if(dis[i][j]<INF) printf("%d ",dis[i][j]);
             else printf("INF ");
         }
         printf("\n");
     }
     */
 }

 int main()
 {
     int i,v,c,r,t,a,b,sum;
     char ch[];
     string s1,s2;
     t=;
     //freopen("data.txt","r",stdin);
     while(scanf("%d %d %d",&n,&c,&r),(n+c+r))
     {
         //printf("%d %d %d\n",n,c,r);
         t++;
         c++;
         M.clear();
         memset(amount,,sizeof(amount));
         n=;
         for(i=;i<=c;i++)
         {
             cin>>s1;
             if(M.count(s1)<=)
             {
                 M[s1]=n++;
             }
             amount[M[s1]]++;
         }
         c=n-;
         memset(dis,,sizeof(dis));
         while(r--)
         {
             cin>>s1;
             if(M.count(s1)<=) M[s1]=n++;
             a=M[s1];
             scanf("%s",ch);
             cin>>s2;
             if(M.count(s2)<=) M[s2]=n++;
             b=M[s2];
             sscanf(ch+,"%d",&v);
             int len=strlen(ch);
             if(ch[len-]=='>')
             {
                 if(dis[a][b]== || dis[a][b]>v) dis[a][b]=v;
             }
             if(ch[]=='<')
             {
                 if(dis[b][a]== || dis[b][a]>v) dis[b][a]=v;
             }
         }
         n--;
         Flyod();
         sum=;
         for(i=;i<=c;i++)
         {
             sum+=(dis[][i]+dis[i][])*amount[i];
         }
         printf("%d. %d\n",t,sum);
     }
     return ;
 }

2923