hdu 2923 Einbahnstrasse

Einbahnstrasse

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1400    Accepted Submission(s): 389

Problem DescriptionEinbahnstra 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.

 

InputYour 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.

 

OutputFor 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 5NewTroy Midvale MetrodaleNewTroy   <-20-> MidvaleMidvale   --50-> BakerlineNewTroy    <-5-- BakerlineMetrodale <-30-> NewTroyMetrodale  --5-> Bakerline0 0 0

 

Sample Output

1. 80

 

Source2008 ANARC
 

Recommendlcy 

題目就是算。你找到一份新工作就是把爛車運回公司。問你完成一天工作後所走的最短路程是多少。

題目不難。關鍵是輸入資料不好處理。

#include <stdio.h>#include<string.h>__int64 dis[150][150];__int64 INF=111111111111;//貌似資料很大於是用int64了。不過同學用int還是過了int n,sum;char city[110][20];void floyd(){    int k,i,j;    for(k=0; k<n; k++)        for(i=0; i<n; i++)            for(j=0; j<n; j++)                if(dis[i][k]+dis[k][j]<dis[i][j])                    dis[i][j]=dis[i][k]+dis[k][j];}int getp(char *c)//建立地點名字和地點號的映射。接收地點名返回地點號//有利於資料處理{    int i;    for(i=0; i<sum; i++)    {        if(strcmp(c,city[i])==0)            return i;    }    strcpy(city[i],c);    sum++;    return i;}int main(){    int a,b,c,r,i,j,left,right,wait[1010];//wait記錄爛車所在的地點號    __int64 d,alld;//d記錄兩地點路的長度.記錄當天總距離    char s[20],e[20],op[10],w[20];//s記錄開始城市名.e記錄結束城市名.op記錄剪頭                                  //和距離。    while(scanf("%d%d%d",&n,&c,&r),n||c||r)    {        t++;//記錄測試數        getchar();        sum=0;        for(i=0; i<n; i++)            for(j=0; j<n; j++)            {                if(i==j)                    dis[i][j]=0;                else                    dis[i][j]=INF;            }        for(i=0; i<c+1; i++)        {            scanf("%s",w);            wait[i]=getp(w);        }        for(i=0; i<r; i++)        {            scanf("%s%s%s",s,op,e);            a=getp(s);            d=0;            left=right=0;            for(j=0; op[j]!='\0'; j++)//記錄道路類型（單雙行道）            {                if(op[j]=='<')                    left=1;//左邊通                if(op[j]=='>')                    right=1;//右邊通                if(op[j]>='0'&&op[j]<='9')                    d=d*10+op[j]-'0';//算出道路長度            }            b=getp(e);            if(left&&d<dis[b][a])                dis[b][a]=d;            if(right&&d<dis[a][b])                dis[a][b]=d;            //printf("%d -> %d is %I64d\n",a,b,dis[a][b]);            //printf("%d -> %d is %I64d\n",b,a,dis[b][a]);        }        floyd();        //for(i=0; i<n; i++)        //for(j=0; j<n; j++)        //printf("%d -> %d is %I64d\n",i,j,dis[i][j]);        alld=0;        for(i=1; i<=c; i++)//每次從公司出發只能處理一輛爛車後必須先回公司        {            alld=alld+dis[wait[0]][wait[i]]+dis[wait[i]][wait[0]];        }        printf("%d. %I64d\n",t,alld);    }    return 0;}/*4 2 5NewTroy Midvale MetrodaleNewTroy   <-20-> MidvaleMidvale   --50-> BakerlineNewTroy    <-5-- BakerlineMetrodale <-30-> NewTroyMetrodale  --5-> Bakerline*/