Minimum Transport Cost

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

Source

Asia 1996, Shanghai (Mainland China)

题意：n座城市，出租车经过编号为i城市要交的费用是cost[i]，给出n*n的邻接矩阵，求多对从s到e的最小花费，并按字典序打印路径。

解析：由于有多对起始点，所以用Floyd算法。但是要稍微处理一下，每次松弛操作的判断条件要改一下，因为总费用要加上出租车经过城市的费用。对于记录字典序最小路径，也是在Floyd里面判断两路径相同时，取字典序小的地点。注意path[i][j]里面保存的是i到j路径上的经过的第一个点，输出的时候，只需要一直回溯输出即可。详见代码

AC代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
using namespace std;
#define INF 1e7 + 2

int n, m;
int d[202][202], v[202];
int cost[202];
int path[202][202];

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++){
                int temp = d[i][k] + d[k][j] + cost[k];     //加上cost[k]
                if(d[i][j] > temp){
                    d[i][j] = temp;
                    path[i][j] = path[i][k];
                }
                else if(d[i][j] == temp){
                    if(path[i][j] > path[i][k])     //取字典序小的
                        path[i][j] = path[i][k];
                }
            }
}

void print(int s, int e){              //打印路径
    printf("%d", s);
    while(e != s){
        printf("-->%d", path[s][e]);
        s = path[s][e];
    }
    puts("");
}

int main(){
    #ifdef sxk
        freopen("in.txt", "r", stdin);
    #endif //sxk

    int x, y, z, s , t;
    while(scanf("%d", &n)!=EOF && n){
        for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++){
                scanf("%d", &d[i][j]);
                if(d[i][j] == -1) d[i][j] = INF;
            }
        for(int i=1; i<=n; i++) scanf("%d", &cost[i]);
        floyd();
        for(int i=0; ; i++){
            scanf("%d%d", &x, &y);
            if(x == -1 && y == -1) break;
            printf("From %d to %d :\n", x, y);
            printf("Path: ");
            print(x, y);
            printf("Total cost : %d\n", d[x][y]);
            puts("");
        }
    }
    return 0;
}

PS：真的学习了，第一次写Floyd的打印路径，还是按字典序，真是太神奇了~~~