hdu 1385 Minimum Transport Cost

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Minimum Transport Cost

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11673    Accepted Submission(s): 3280

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

50 3 22 -1 43 0 5 -1 -122 5 0 9 20-1 -1 9 0 44 -1 20 4 05 17 8 3 11 33 52 4-1 -10

 

Sample Output

From 1 to 3 :Path: 1-->5-->4-->3Total cost : 21From 3 to 5 :Path: 3-->4-->5Total cost : 16From 2 to 4 :Path: 2-->1-->5-->4Total cost : 17

题意：先给你一张你n * n的图，代表城市间的距离，然后，给出n个tax的费用，然后很多询问，问你a到b的最少费用。。。。并且打印路径（字典序）。。。

最短路（Floyd）+打印路径

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int cost[10005][10005];int dp[10005][10005];int tax[10005];const int inf=0x3f3f3f3f;int main(){int n;while(cin>>n,n){for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){cin>>cost[i][j];if(cost[i][j]==-1)cost[i][j]=inf;dp[i][j]=j;//初始化}}for(int i=1;i<=n;i++)cin>>tax[i];for(int k=1;k<=n;k++){for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){if(cost[i][j]>cost[i][k]+cost[k][j]+tax[k]){cost[i][j]=cost[i][k]+cost[k][j]+tax[k];dp[i][j]=dp[i][k];}else{if(cost[i][j]==cost[i][k]+cost[k][j]+tax[k]&&dp[i][j]>dp[i][k])dp[i][j]=dp[i][k];}}}}int a,b;while(cin>>a>>b){if(a==-1&&b==-1)break; int u=a,v=b; printf("From %d to %d :\nPath: %d",a,b,a); while(u!=v) { printf("-->%d",dp[u][v]); u=dp[u][v]; } printf("\nTotal cost : %d\n\n",cost[a][b]);}}return 0;}

参考博客

https://www.cnblogs.com/qiufeihai/archive/2012/09/05/2672015.html