hdu1358 Minimum Transport Cost 按字典序输出最短路径hdu1385

最新推荐文章于 2025-03-27 16:42:40 发布

最新推荐文章于 2025-03-27 16:42:40 发布 · 195 阅读

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本文介绍了一个经典的图论问题——寻找两个城市间具有最小成本的货物运输路径，包括运输费和过境税。通过Dijkstra算法的改进版实现，并考虑了路径字典序最小的要求。

Minimum Transport Cost

Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3923Accepted Submission(s): 989

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

Source

Asia 1996, Shanghai (Mainland China)

#include<stdio.h>
#include<string.h>
#define size 1000
int val[size],map[size][size],used[size],min[size],n,start[size],end[size],flag,pre[size];
char s1[1300],s2[1300];
int ok(int n1,int n2,int s)//比较字典序大小
{
	int i,k;
    i=0;
    s1[i++]=n1+'0';
    k=pre[n1];
    while(k!=s)
    {
		s1[i++]=k+'0';
		k=pre[k];
    }
    s1[i++]=s+'0';
    s1[i]='\0';
    strrev(s1);//把字符串翻转过来
    i=0;
    s2[i++]=n1+'0';
    k=n2;//因为依然是n1(j)为最后一个 通过n2(pos)架桥
    while(k!=s)
    {
		s2[i++]=k+'0';
		k=pre[k];
    }
    s2[i++]=s+'0';
    s2[i]='\0';
    strrev(s2);
    if(strcmp(s2,s1)<0)
		return 1;
    else  return 0;
}
void find_short(int s)
{
	int i,j,mm,pos;
	memset(used,0,sizeof(used));
	for(i=1;i<=n;i++)
	{
		min[i]=map[s][i];
		if(map[s][i]!=999999999) //这一步居然忘记了 阿弥陀佛  记住这个千万不能少
			pre[i]=s;
	}
	min[s]=0;used[s]=1;
	for(i=2;i<=n;i++)
	{
		mm=999999999;
		for(j=1;j<=n;j++)
		{
			if(!used[j]&&mm>min[j])
			{
				pos=j;
				mm=min[j];
			}
		}
		if(mm==999999999) break;
		used[pos]=1;
		for(j=1;j<=n;j++)
			if(!used[j]&&min[pos]+map[pos][j]+val[pos]<min[j])
			{
				min[j]=min[pos]+map[pos][j]+val[pos];
				pre[j]=pos;
			}
			else
				if(!used[j]&&min[pos]+map[pos][j]+val[pos]==min[j])
					if(ok(j,pos,s)) pre[j]=pos;//如果pos之前的字典序较小那么更新
	}
}
void print(int s,int e)
{
	int a[size],i,k;
	k=0;
	a[k++]=e;
	i=e;
	while(s!=i)
	{
		i=pre[i];
		a[k++]=i;
	}
	printf("Path: ");
	for(i=k-1;i>0;i--)
		printf("%d-->",a[i]);
	printf("%d\n",a[0]);
}
int main()
{
	int i,j,s,e,num;
	while(scanf("%d",&n)&&n!=0)
	{
		num=0;
		for(i=1;i<=n;i++)
			for(j=1;j<=n;j++)
			{
				scanf("%d",&map[i][j]);
				if(map[i][j]==-1) map[i][j]=999999999;
			}
			for(i=1;i<=n;i++)
				scanf("%d",&val[i]);
			while(1)
			{
				scanf("%d %d",&s,&e);
				if(s==-1&&e==-1) break;
				start[num]=s;
				end[num++]=e;
			}
			for(i=0;i<num;i++)
			{
				flag=0;
				find_short(start[i]);
				printf("From %d to %d :\n",start[i],end[i]);
				if(start[i]!=end[i])
					print(start[i],end[i]);
				else printf("Path: %d\n",start[i]);
				printf("Total cost : %d\n",min[end[i]]);
				printf("\n");
			}
	}
	return 0;
}
/*记录路径思路就是加入一个pre[]数组 时时更新每个点前面的点是什么  另外不要忘记从始点出发到i的所有的pre都要赋值初始点
本题要求按字典序找出最短的路径  一开始我感到很恐惧 不知道怎么做  其实心里没有必要那么害怕 只要尝试着一点一点的去做
不要不敢尝试就退却   把大问题分解成一个一个小问题   硬着头皮也要上  (看了别人代码才A了)*/