zoukankan      html  css  js  c++  java
  • Minimum Transport Cost Floyd 输出最短路

    These are N cities in Spring country. Between each pair of cities there may be one transportation track or none. Now there is some cargo that should be delivered from one city to another. The transportation fee consists of two parts: 
    The cost of the transportation on the path between these cities, and 

    a certain tax which will be charged whenever any cargo passing through one city, except for the source and the destination cities. 

    You must write a program to find the route which has the minimum cost. 

    InputFirst is N, number of cities. N = 0 indicates the end of input. 

    The data of path cost, city tax, source and destination cities are given in the input, which is of the form: 

    a11 a12 ... a1N 
    a21 a22 ... a2N 
    ............... 
    aN1 aN2 ... aNN 
    b1 b2 ... bN 

    c d 
    e f 
    ... 
    g h 

    where aij is the transport cost from city i to city j, aij = -1 indicates there is no direct path between city i and city j. bi represents the tax of passing through city i. And the cargo is to be delivered from city c to city d, city e to city f, ..., and g = h = -1. You must output the sequence of cities passed by and the total cost which is of the form: 
    OutputFrom c to d : 
    Path: c-->c1-->......-->ck-->d 
    Total cost : ...... 
    ...... 

    From e to f : 
    Path: e-->e1-->..........-->ek-->f 
    Total cost : ...... 

    Note: if there are more minimal paths, output the lexically smallest one. Print a blank line after each test case. 

    Sample Input

    5
    0 3 22 -1 4
    3 0 5 -1 -1
    22 5 0 9 20
    -1 -1 9 0 4
    4 -1 20 4 0
    5 17 8 3 1
    1 3
    3 5
    2 4
    -1 -1
    0

    Sample Output

    From 1 to 3 :
    Path: 1-->5-->4-->3
    Total cost : 21
    
    From 3 to 5 :
    Path: 3-->4-->5
    Total cost : 16
    
    From 2 to 4 :
    Path: 2-->1-->5-->4
    Total cost : 17

    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<cstring>
    #include<sstream>
    #include<algorithm>
    #include<queue>
    #include<deque>
    #include<iomanip>
    #include<vector>
    #include<cmath>
    #include<map>
    #include<stack>
    #include<set>
    #include<fstream>
    #include<memory>
    #include<list>
    #include<string>
    using namespace std;
    typedef long long LL;
    typedef unsigned long long ULL;
    #define MAXN 103
    #define L 31
    #define INF 1000000009
    #define eps 0.00000001
    int g[MAXN][MAXN],path[MAXN][MAXN], n, v[MAXN];//path[i][j] 表示路径i->j上 i之后的第一个结点
    void Floyd()
    {
        for (int k = 1; k <= n; k++)
        {
            for (int i = 1; i <= n; i++)
            {
                for (int j = 1; j <= n; j++)
                {
                    if (g[i][j] > g[i][k] + g[k][j] + v[k])
                    {
                        g[i][j] = g[i][k] + g[k][j] + v[k];
                        path[i][j] = path[i][k];//将路径结点缩小范围确定
                    }
                    else if (g[i][j] == g[i][k] + g[k][j] + v[k])
                    {
                        if (path[i][j] > path[i][k])
                        {
                            path[i][j] = path[i][k];//比较前面的字典序
                        }
                    }
                }
            }
        }
    }
    void Print(int u, int v)//递归 从i->j 可以分解为 i->path[i][j]->path[path[i][j]][j]->,,,,,j
    {
        if (u == v)
        {
            printf("%d
    ", u);
            return;
        }
        int k = path[u][v];
        printf("%d-->", u);
        Print(k, v);
    }
    int main()
    {
        while (scanf("%d", &n), n)
        {
            for (int i = 1; i <= n; i++)
            {
                for (int j = 1; j <= n; j++)
                {
                    scanf("%d", &g[i][j]);
                    path[i][j] = j;
                    if (g[i][j] == -1) g[i][j] = INF;
                }
            }
            for (int i = 1; i <= n; i++)
                scanf("%d", &v[i]);
            int f, t;
            Floyd();
            while (scanf("%d%d", &f, &t), f != -1 && t != -1)
            {
                printf("From %d to %d :
    Path: ", f, t);
                Print(f, t);
                printf("Total cost : %d
    
    ", g[f][t]);
            }
        }
    }
  • 相关阅读:
    胖子哥的大数据之路(12)-三张图告诉你大数据安全方案设计
    Building the Unstructured Data Warehouse: Architecture, Analysis, and Design
    Hadoop专业解决方案-第3章:MapReduce处理数据
    胖子哥的大数据之路(11)-我看Intel&&Cloudera的合作
    胖子哥的大数据之路(10)- 基于Hive构建数据仓库实例
    胖子哥的大数据之路(9)-数据仓库金融行业数据逻辑模型FS-LDM
    胖子哥的大数据之路(8)- 数据仓库命名规范
    胖子哥的大数据之路(7)- 传统企业切入核心or外围
    dos 打开计算机管理
    C# SQLite数据库
  • 原文地址:https://www.cnblogs.com/joeylee97/p/7343966.html
Copyright © 2011-2022 走看看