Minimum Transport Cost
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2649 Accepted Submission(s): 651
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
这题是一道很好的题目:输出最短路路径,方法:记录每一点的前一个点为多少,用floyd的话就是要二维了。
代码:
#include <iostream>
#include <stdio.h>
#include <memory.h>
#include <queue>
using namespace std;
const int INF = 99999999;
const int N = 505;
int map[N][N], tax[N], path[N][N];
int n;
void init()
{
int i, j;
for(i = 1; i <= n; i++)
for(j = 1; j <= n; j++)
if(i == j) map[i][j] = 0;
else map[i][j] = INF;
}
void input()
{
int i, j, k;
for(i = 1; i <= n; i++)
for(j = 1; j <= n; j++)
{
scanf("%d", &k);
if(k != -1) map[i][j] = k;
path[i][j] = j;
}
for(i = 1; i <= n; i++)
scanf("%d", &tax[i]);
}
void floyd()
{
int i, j, k, len;
for(k = 1; k <= n; k++)
{
for(i = 1; i <= n; i++)
{
for(j = 1; j <= n; j++)
{
len = map[i][k] + map[k][j] + tax[k];
if(map[i][j] > len)
{
map[i][j] = len;
path[i][j] = path[i][k]; //标记到该点的前一个点
}
else if(len == map[i][j]) //若距离相同
{
if(path[i][j] > path[i][k]) //判断是否为字典顺序
path[i][j] = path[i][k];
}
}
}
}
}
void output()
{
int i, j, k;
while(scanf("%d %d", &i, &j))
{
if(i == -1 && j == -1) break;
printf("From %d to %d :\n", i, j);
printf("Path: %d", i);
k = i;
while(k != j) //输出路径从起点直至终点
{
printf("-->%d", path[k][j]);
k = path[k][j];
}
printf("\n");
printf("Total cost : %d\n\n", map[i][j]);
}
}
int main()
{
while(scanf("%d", &n), n)
{
init();
input();
floyd();
output();
}
return 0;
}
分享到:
相关推荐
最短路(HDU-2544).rar
HDU最全ac代码
HDU的1250,主要是利用高精度加法,但是代码有点繁琐,效率不是很高
杭电ACMhdu1163
HDU1059的代码
hdu1001解题报告
hdu 1574 passed sorce
HDU的一题........HDU DP动态规
hdu2101AC代码
hdu acm 教案 搜索入门 hdu acm 教案 搜索入门
搜索 dfs 解题代码 hdu1241
最短路径 贪心实现代码 hdu 最短路 dijkstra 算法
hdu 5007 Post Robot 字符串枚举。 暴力一下就可以了。
hdu acm 教案 动态规划(1) hdu acm 教案 动态规划(1)
hdu 1166线段树代码
ACM HDU题目分类,我自己总结的大概只有十来个吧
自己做的HDU ACM已经AC的题目
hdu动态规划算法集锦
hdu题目分类
HDU图论题目分类