HDU1385Minimum Transport Cost（Flody+bfs路径输出）

题目来源：
http://acm.hdu.edu.cn/showproblem.php?pid=1385

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
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

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

题意：

    给你一个无向图,现在要你输出特定起点到终点的最短距离以及字典序最小的路径.

思路：

Floyd+bfs路径输出

参考代码：

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdio>
#define inf 0x3f3f3f3f
#define N 1010
using namespace std;
int n,cost[N],dist[N][N],path[N][N];//dist[i][j]表示i到j的最短距离,path[i][j]保存了从i到j路径的第一个点(除i以外)
void Floyd()
{
    for(int i=1; i<=n; i++)
        for(int j=1; j<=n; j++)
            path[i][j]=j;
    for(int k=1; k<=n; k++)
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
                if(dist[i][k]<inf && dist[k][j]<inf)
                {
                    if(dist[i][j] > dist[i][k]+dist[k][j]+cost[k])
                    {
                        dist[i][j] = dist[i][k]+dist[k][j]+cost[k];
                        path[i][j] = path[i][k];
                    }
                    else if(dist[i][j] == dist[i][k]+dist[k][j]+cost[k] && path[i][j]>path[i][k])
                    {
                        path[i][j]=path[i][k];
                    }
                }
}
void bfs(int u,int v)
{
    printf("From %d to %d :\n",u,v);
    if(u!=v)
    {
        printf("Path: %d",u);
        int now = path[u][v];
        while(1)
        {
            printf("-->%d",now);
            if(now==v)
            {
                printf("\n");
                break;
            }
            now = path[now][v];
        }
    }
    else//注意U==V的特殊情况
    {
        printf("Path: %d\n",u);
    }
    printf("Total cost : %d\n\n",dist[u][v]);
}
int main()
{
    while(~scanf("%d",&n))
    {
        if(!n)break;
        memset(dist,0,sizeof(dist));
		memset(path,0,sizeof(path));
		memset(cost,0,sizeof(cost));
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++)
            {
                scanf("%d",&dist[i][j]);
                if(dist[i][j]==-1) dist[i][j]=inf;
            }
        for(int i=1; i<=n; i++)
            scanf("%d",&cost[i]);
        Floyd();
        while(1)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            if(u==-1&&v==-1)
            {
                break;
            }
            bfs(u,v);
        }
    }
    return 0;
}