ZOJ 1456 Minimum Transport Cost（Floyd算法求解最短路径并输出最小字典序路径）

题目链接：

https://vjudge.net/problem/ZOJ-1456

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

/*
题意描述
    给出路径花费矩阵，路途中经过结点的过路费，问从a到b的短路径是多少，并输出路径，如果最小花费相同，输出路径字典序最小的那条
解题思路
    使用Floyd，每次检测到某个节点能够松弛某条路径，就更新直接后继，二维数组记录路径的方法。 
*/ 
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <vector>
#include <cstring>
using namespace std;

const int inf = 99999999;
const int maxn = 1010;
int e[maxn][maxn], suc[maxn][maxn], b[maxn];
int n;
void Floyd();
 
int main()
{
    while(scanf("%d", &n) == 1 && n) {
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= n; j++) {
                scanf("%d", &e[i][j]);
                if(e[i][j] == -1)
                    e[i][j] = inf;
                suc[i][j] = j;//从i到j的直接后继初始化为j 
            }
        }
        for(int i = 1; i <= n; i++) {
            scanf("%d", &b[i]);
        }
        Floyd(); 
        int u, v;
        while(scanf("%d%d", &u, &v) == 2 && u + v != -2) {
            printf("From %d to %d :
Path: %d", u, v, u);
            int tp = u;
            while(tp != v) {//一直输出u到v的直接后继，直到后继就是v 
                printf("-->%d", suc[tp][v]);
                tp = suc[tp][v];
            } 
            printf("
Total cost : %d

", e[u][v]);
        }
    }
    return 0;
}

void Floyd() {
    for(int k = 1; k <= n; k++) {
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= n; j++) {
                if(e[i][j] > e[i][k] + e[k][j] + b[k]) {
                    e[i][j] = e[i][k] + e[k][j] + b[k];
                    suc[i][j] = suc[i][k];//如果k点能够使从i到j的路径松弛，那么就将i到j的直接后继更新为从i到k的直接后继 
                } else if(e[i][j] == e[i][k] + e[k][j] + b[k] && suc[i][j] > suc[i][k]) 
                    suc[i][j] = suc[i][k];
            }
        }
    }
}