HDU 1385 Minimum Transport Cost

 

首先，这道题考的的内容有三点：最短路+路径标记+字典序输出

最短路这里用Floyd直接就搞出来了，二路径标记和字典序只要少做一下处理便好

 

Minimum Transport Cost

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5245    Accepted Submission(s): 1316

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

 

#include<iostream>
#include<string>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;

#define N 1010
#define MAX 0x3f3f3f3f
int v, dis[N][N], tax[N], path[N][N];

void Floyd()
{
    for(int i = 1; i <= v; ++i)
        for(int j = 1; j <= v; ++j)
            path[i][j] = j; //i从j点可到达j
    for(int k = 1; k <= v; ++k) //Floyd算法
        for(int i = 1; i <= v; ++i)
            for(int j = 1; j <= v; ++j)
            {
                int temp = dis[i][k] + dis[k][j] + tax[k];
                if(temp < dis[i][j]) //松弛，更新最短路
                {
                    dis[i][j] = temp;
                    path[i][j] = path[i][k]; //i经k到达j比i经j到达j短
                }
                else if(temp == dis[i][j] && path[i][j] > path[i][k]) //寻找字典序更小的路径
                    path[i][j] = path[i][k];
            }
}

int main()
{
    int start, end;
    while(scanf("%d", &v) != EOF && v)
    {
        for(int i = 1; i <= v; ++i)
            for(int j = 1; j <= v; ++j)
            {
                scanf("%d", &dis[i][j]);
                dis[i][j] = (dis[i][j] == -1 ? MAX : dis[i][j]); //-1无路赋值极大
            }
        for(int i = 1; i <= v; ++i) //税
            scanf("%d", &tax[i]);
        Floyd();
        while(scanf("%d%d", &start, &end) != EOF && start != -1 && end != -1)
        {
            printf("From %d to %d :\n", start, end);
            printf("Path: %d", start);
            int res = start;
            while(res != end) //最短路径
            {
                printf("-->%d", path[res][end]);
                res = path[res][end];
            }
            printf("\nTotal cost : %d\n\n", dis[start][end]); //最短路权值
        }
    }
    return 0;
}