zoukankan      html  css  js  c++  java
  • Floyd求字典序最小的路径

    hdu1384

    Minimum Transport Cost

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 7011    Accepted Submission(s): 1793


    Problem Description
    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.
     

    Input
    First 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:
     

    Output
    From 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
    #include"stdio.h"
    #include"string.h"
    #include"iostream"
    #define M 111
    #define inf 99999999
    #define eps 1e-9
    #include"math.h"
    using namespace std;
    int G[M][M],dis[M][M],path[M][M],n,b[M];
    void floyd()
    {
        int i,j,k;
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=n;j++)
            {
                dis[i][j]=G[i][j];
                path[i][j]=j;
            }
        }
        for(k=1;k<=n;k++)
        {
            for(i=1;i<=n;i++)
            {
                for(j=1;j<=n;j++)
                {
                    if(dis[i][j]>dis[i][k]+dis[k][j]+b[k])
                    {
                        dis[i][j]=dis[i][k]+dis[k][j]+b[k];
                        path[i][j]=path[i][k];
                    }
                    else if(dis[i][j]==dis[i][k]+dis[k][j]+b[k])
                    {
                        if(path[i][j]>path[i][k]&&i!=k)
                            path[i][j]=path[i][k];
                    }
                }
            }
        }
    }
    void solve(int i,int j)
    {
        printf("From %d to %d :
    ",i,j);
        printf("Path: ");
        int k=i;
        printf("%d",i);
        while(k!=j)
        {
            printf("-->%d",path[k][j]);
            k=path[k][j];
        }
        printf("
    ");
        printf("Total cost : %d
    
    ",dis[i][j]);
    }
    int main()
    {
        int i,j;
        while(scanf("%d",&n),n)
        {
            for(i=1;i<=n;i++)
            {
                for(j=1;j<=n;j++)
                {
                    G[i][j]=inf;
                }
                G[i][i]=0;
            }
            for(i=1;i<=n;i++)
            {
                for(j=1;j<=n;j++)
                {
                    int a;
                    scanf("%d",&a);
                    if(a!=-1)
                        G[i][j]=a;
                }
            }
            for(i=1;i<=n;i++)
                scanf("%d",&b[i]);
            floyd();
            int s,e;
            while(scanf("%d%d",&s,&e),s!=-1||e!=-1)
            {
                solve(s,e);
            }
        }
        return 0;
    }
    

    3 52 4-1 -10
     

    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
    程序:


  • 相关阅读:
    HDU 3605 Escape 最大流
    HDU 3416 Marriage Match IV (最短路径&&最大流)
    洛谷1508 简单记忆化搜索
    洛谷1880 区间dp+记忆化搜索 合并石子
    洛谷1063 +区间dp(经典问题)
    洛谷1074 靶状数独dfs 排序、记录、搜索
    hdu3368 dfs 下棋
    hdu1258 dfs 给一个指定的target数和一个数列,要求不重复选择其中的数使得和为target并打印,结果不可重复。
    hdu1181 dfs 字符串首尾可拼接,问是否可寻找到一条字串路径使得首尾分别是‘b’和‘m’,简单的搜索+回溯
    hdu1078 dfs+dp(记忆化搜索)搜索一条递增路径,路径和最大,起点是(0,0)
  • 原文地址:https://www.cnblogs.com/mypsq/p/4348228.html
Copyright © 2011-2022 走看看