Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Brightness of Brain Contest Problem H Time limit: 1 second Memory: 32 MB

Cheapest way

The Problem

``Moutushi Enterprise'' a leading transport company of Bangladesh is quite anxious about their current business status. For some recent months they aren't earning profitable money as they did before. They investigated over it and found that it is the change in the tax policy that caused them so. For this reason they want to reroute their ways and charge the passengers on the cost of that route.

The cost a bus has to bear is the cost of the oil it needs for that journey, along with some money paid on each station it touches. The profit is ten percent of the cost. Then the amount is divided by the total number of sits in the bus and the seat fare is then determined.

You are given the description of the city map, you are to determine the cheapest way of the map and the seat fare each passenger have to bear.
 

The Input

The first line of the input indicates the number of map to process. Each map contains three sections. Section A contains the stations that the map contains. First line of this section is an integer ``nStation'' (0<`nStation'<20) indicating number of stations the path have. Next `nStation' line contains station name and the money that have to pay to pass through this station. Next section starts with an integer ``nPath'' (0<`nPath'<20)indicating number of paths that connects two stations. Next `nPath' contains ``station1 station2 distance'', describing `station1' has a path to `station2' with distance equal to `distance' in kilometer and vice versa. Each kilometer costs an oil consumption of 2 taka (unit of money in Bangladesh). The third section consists of a integer ``nQuery'' (0<`nQuery'<10)with `nQuery' line of input. Each line have a query to follow in the format ``station1 station2 passenger'', asking to find the best path from station1 to station2 and what will be the cost for each passenger for a bus with `passenger' amount of seat. It is guaranteed that each map description is valid, and the query has a valid and unique path.

The Output

For each map print ``Map #X'', where `X' is the map number starting from 1. Then for each query print ``Query #Y'', where `Y' is the query number for each map starting from 1. Then for each query print the suitable path followed by the fare each passenger have to bear (rounded to the second digit after the decimal point) as shown below.

Sample Input

2

4
mirpur12 5
farmgate 8
gulistan 10
newmarket 5
4
mirpur12 farmgate 12
mirpur12 newmarket 20
farmgate gulistan 10
newmarket gulistan 8
2
mirpur12 gulistan 30
mirpur12 newmarket 30

3
uttara 2
farmgate 8
gulistan 10
2
uttara farmgate 35
farmgate gulistan 10
1
uttara gulistan 30

Sample Output

Map #1
Query #1
mirpur12 farmgate gulistan
Each passenger has to pay : 2.46 taka
Query #2
mirpur12 newmarket
Each passenger has to pay : 1.83 taka

Map #2
Query #1
uttara farmgate gulistan
Each passenger has to pay : 4.03 taka

---

Suman Mahbub
Created: 11-10-2002
Updated: 14-12-2002, 20-01-2003

---

題目描述：

每經過一個點就會收費，類似收費亭的概念，每條路都是雙向的，每一公里消耗 2 單位。
最後將成本乘上 1.1，將這些費用交給使用的所有乘客去平攤。

題目解法：

floyd-warshall 解法，由於每過一個點就會收費，直接對每條邊的權重多上端點權重。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <map>
#include <algorithm>
using namespace std;

int g[105][105], next[105][105];
int main() {
    int testcase, cases = 0;
    int i, j, k, n, m;
    int x, y, v;
    char s[105];
    char sname[105][105];
    int a[105];
    scanf("%d", &testcase);
    while(testcase--) {
        scanf("%d", &n);
        map<string, int> R;
        for(i = 0; i < n; i++) {
            scanf("%s %d", sname[i], &a[i]);
            R[sname[i]] = i;
        }
        scanf("%d", &m);
        for(i = 0; i < n; i++) {
            for(j = 0; j < n; j++)
                g[i][j] = 0xfffffff, next[i][j] = j;
            g[i][i] = 0;
        }
        for(i = 0; i < m; i++) {
            scanf("%s", s);
            x = R[s];
            scanf("%s", s);
            y = R[s];
            scanf("%d", &v);
            v *= 2;
            g[x][y] = min(g[x][y], v);
            g[y][x] = min(g[y][x], v);
        }
        for(i = 0; i < n; i++) {
            for(j = 0; j < n; j++) {
                g[i][j] += a[j];
            }
            g[i][i] = 0;
        }
        for(k = 0; k < n; k++)
            for(i = 0; i < n; i++)
                for(j = 0; j < n; j++)
                    if(g[i][j] > g[i][k] + g[k][j]) {
                        g[i][j] = g[i][k] + g[k][j];
                        next[i][j] = next[i][k];
                    }
        scanf("%d", &m);
        printf("Map #%d\n", ++cases);
        int qcases = 0;
        while(m--) {
            scanf("%s", s);
            x = R[s];
            scanf("%s", s);
            y = R[s];
            scanf("%d", &v);
            printf("Query #%d\n", ++qcases);
            double pay = (g[x][y] + a[x]) * 1.1 / v;
            printf("%s", sname[x]);
            x = next[x][y];
            while(x != y) {
                printf(" %s", sname[x]);
                x = next[x][y];
            }
            printf(" %s", sname[x]);
            puts("");
            printf("Each passenger has to pay : %.2lf taka\n", pay);
        }
        if(testcase)
            puts("");
    }
    return 0;
}