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HDU 2923 Einbahnstrasse（最短路 Floyd）

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HDU 2923 Einbahnstrasse（最短路 Floyd）

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-> Midvale
Midvale   --50-> Bakerline
NewTroy    <-5-- Bakerline
Metrodale <-30-> NewTroy
Metrodale  --5-> Bakerline
0 0 0

Sample Output

1. 80

題意：給出三個數 n， c， r

n 個地點（包括公司的車庫），c 表示c輛車拋錨的地點， r 條道路

第二行給出 c+1 個地點，第一個為車庫地點，其余的 c 個為車的地點。

接下來的 r 行表示 r 條有向的道路，

s1 -- d -> s2 表示 s1到s2 的長度為 d

s1 <- d -- s2 表示 s2到s1 的長度為 d

s1 <- d -> s2 表示 s1到s2為雙向邊，且長度為 d

拖車從車庫出發到每個地點，在該地點拖回拋錨的車子。

一輛拖車一次只能拖回一輛車子。

注意：同一個地點可能有多輛拋錨的車子

求總的路徑長度。

還要輸出 case number

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include
using namespace std;

const double PI = acos(-1.0);
const double e = 2.718281828459;
const double eps = 1e-8;
const int MAXN = 1010;
const int INF = 1<<29;
char str[1010][1100];
int dist[110][1100];
char s1[110];
char s2[110];
mapm;
int n;

void Floyd()
{
    for(int k = 1; k <= n; k++)
    {
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= n; j++)
                if(dist[i][j]>dist[i][k]+dist[k][j])
                    dist[i][j] = dist[i][k]+dist[k][j];
        }
    }
}

int main()
{
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int c, r;
    int num = 1;
    while(cin>>n>>c>>r)
    {
        if(!n && !c && !r)
            break;
        for(int i = 0; i <= n; i++)
        {
            for(int j = 0; j <= n; j++)
                dist[i][j] = (i==j?0:INF);
        }
        for(int i = 0; i <= c; i++)
        {
            cin>>str[i];
        }
        int d, x, y;
        char from, to;
        int cnt = 1;
        m.clear(); // 注意清空，wa 了很多次
        for(int i = 0; i < r; i++)
        {
            scanf("%s %c-%d-%c %s", s1, &from, &d, &to, s2);
            if(!m[s1])
                m[s1] = cnt++;
            if(!m[s2])
                m[s2] = cnt++;
            x = m[s1];
            y = m[s2];
            if(from=='<' && d' && d