Bus System
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6166 Accepted Submission(s): 1580
Problem Description
Because of the huge population of China, public transportation is very important. Bus is an important transportation method in traditional public transportation system. And it’s still playing an important role even now.
The bus system of City X is quite strange. Unlike other city’s system, the cost of ticket is calculated based on the distance between the two stations. Here is a list which describes the relationship between the distance and the cost.
Your neighbor is a person who is a really miser. He asked you to help him to calculate the minimum cost between the two stations he listed. Can you solve this problem for him?
To simplify this problem, you can assume that all the stations are located on a straight line. We use x-coordinates to describe the stations’ positions.
Input
The input consists of several test cases. There is a single number above all, the number of cases. There are no more than 20 cases.
Each case contains eight integers on the first line, which are L1, L2, L3, L4, C1, C2, C3, C4, each number is non-negative and not larger than 1,000,000,000. You can also assume that L1<=L2<=L3<=L4.
Two integers, n and m, are given next, representing the number of the stations and questions. Each of the next n lines contains one integer, representing the x-coordinate of the ith station. Each of the next m lines contains two integers, representing the start point and the destination.
In all of the questions, the start point will be different from the destination.
For each case,2<=N<=100,0<=M<=500, each x-coordinate is between -1,000,000,000 and 1,000,000,000, and no two x-coordinates will have the same value.
Output
For each question, if the two stations are attainable, print the minimum cost between them. Otherwise, print “Station X and station Y are not attainable.” Use the format in the sample.
Sample Input
2
1 2 3 4 1 3 5 7
4 2
1
2
3
4
1 4
4 1
1 2 3 4 1 3 5 7
4 1
1
2
3
10
1 4
Sample Output
Case 1:
The minimum cost between station 1 and station 4 is 3.
The minimum cost between station 4 and station 1 is 3.
Case 2:
Station 1 and station 4 are not attainable.
题意：坐公交车不同的距离，所需要的花费是不一样的，给你8个数，前4个代表不同的距离，后4个代表的是相应的花费，若距离不在所规定的范围，则不能到达吧。再输入n和m，分别表示在坐标上有n个点，代表不同的坐标值，m则代表将要询问的最短路的次数了。
解题思路：就是一道dijkstra或floyd的模板题，但要注意数据的大小，要用int64位来计算，而且定义的最大值的是非常大的数。一开始我定义为2^32-1，但结果WA了，后面改了就可以了。
Dijkstra贴代码：
#include <stdio.h>
#define maxn 1e18
__int64 map[105][105], dis[105];
int visited[105];
void Dijkstra(int start, int n)
{
__int64 mind;
int pre = start;
for(int i = 1; i<=n; i++)
{
dis[i] = map[start][i];
visited[i] = 0;
}
visited[start] = 1;
for(int i = 1; i<=n; i++)
{
mind = maxn;
for(int j = 1; j<=n; j++)
{
if(!visited[j] && mind > dis[j])
{
mind = dis[j];
pre = j;
}
}
visited[pre] = 1;
for(int j = 1; j<=n; j++)
{
if(!visited[j] && dis[j] > dis[pre]+map[pre][j])
dis[j] = dis[pre]+map[pre][j];
}
}
}
int main()
{
__int64 x, N[105];
int t = 1, T;
int n, m;
int start, finish;
__int64 L1, L2, L3, L4, C1, C2, C3, C4;
scanf("%d", &T);
while(T--)
{
scanf("%I64d%I64d%I64d%I64d%I64d%I64d%I64d%I64d", &L1, &L2, &L3, &L4, &C1, &C2, &C3, &C4);
scanf("%d%d", &n, &m);
for(int i = 1; i<=n; i++)
scanf("%I64d", &N[i]);
for(int i = 1; i<=n; i++)
{
for(int j = 1; j<=n; j++)
{
if(N[i] > N[j])
x = N[i] - N[j];
else
x = N[j] - N[i];
if(x>0 && x<=L1)
map[i][j] = C1;
else if(x>L1 && x<=L2)
map[i][j] = C2;
else if(x>L2 && x<=L3)
map[i][j] = C3;
else if(x>L3 && x<=L4)
map[i][j] = C4;
else
map[i][j] = maxn;
if(i == j)
map[i][j] = 0;
}
}
printf("Case %d:\n", t++);
while(m--)
{
scanf("%d%d", &start, &finish);
Dijkstra(start, n);
if(visited[finish])
printf("The minimum cost between station %d and station %d is %I64d.\n", start, finish, dis[finish]);
else
printf("Station %d and station %d are not attainable.\n", start, finish);
}
}
return 0;
}
Floyd算法的代码：
#include <stdio.h>
#define maxn 1e18
__int64 map[105][105], dis[105];
void Floyd(int n)
{
for(int k = 1; k<=n; k++)
{
for(int i = 1; i<=n; i++)
{
for(int j = 1; j<=n; j++)
{
if(map[i][j] > map[i][k]+map[k][j])
map[i][j] = map[i][k]+map[k][j];
}
}
}
}
int main()
{
__int64 x, N[105];
int t = 1, T;
int n, m;
int start, finish;
__int64 L1, L2, L3, L4, C1, C2, C3, C4;
scanf("%d", &T);
while(T--)
{
scanf("%I64d%I64d%I64d%I64d%I64d%I64d%I64d%I64d", &L1, &L2, &L3, &L4, &C1, &C2, &C3, &C4);
scanf("%d%d", &n, &m);
for(int i = 1; i<=n; i++)
scanf("%I64d", &N[i]);
for(int i = 1; i<=n; i++)
{
for(int j = 1; j<=n; j++)
{
if(N[i] > N[j])
x = N[i] - N[j];
else
x = N[j] - N[i];
if(x>0 && x<=L1)
map[i][j] = C1;
else if(x>L1 && x<=L2)
map[i][j] = C2;
else if(x>L2 && x<=L3)
map[i][j] = C3;
else if(x>L3 && x<=L4)
map[i][j] = C4;
else
map[i][j] = maxn;
if(i == j)
map[i][j] = 0;
}
}
Floyd(n);
printf("Case %d:\n", t++);
while(m--)
{
scanf("%d%d", &start, &finish);
if(map[start][finish] != maxn)
printf("The minimum cost between station %d and station %d is %I64d.\n", start, finish, map[start][finish]);
else
printf("Station %d and station %d are not attainable.\n", start, finish);
}
}
return 0;
}
hdu 1960 Bus System