HDU 2680  Choose the best route

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Choose the best route

Time Limit: 2000/1000 MS(Java/Others)    Memory Limit:32768/32768 K (Java/Others)
Total Submission(s): 1779   Accepted Submission(s):548


Problem Description
One day , Kiki wants to visit one of her friends. As she isliable to carsickness , she wants to arrive at her friend’s home assoon as possible . Now give you a map of the city’s traffic route,and the stations which are near Kiki’s home so that she can take.You may suppose Kiki can change the bus at any station. Please findout the least time Kiki needs to spend. To make it easy, if thecity have n bus stations ,the stations will been expressed as aninteger 1,2,3…n.
 

Input
There are several test cases. 
Each case begins with three integers n, m ands,(n<1000,m<20000,1=<s<=n)n stands for the number of bus stations in this city and m standsfor the number of directed ways between bus stations .(Maybe thereare several ways between two bus stations .) s stands for the busstation that near Kiki’s friend’s home.
Then follow m lines ,each line contains three integers p , q ,t (0<t<=1000). means from station pto station q there is a way and it will costs t minutes .
Then a line with an integerw(0<w<n), means the number ofstations Kiki can take at the beginning. Then follows w integersstands for these stations.
 

Output
The output contains one line for each data set : the leasttime Kiki needs to spend ,if it’s impossible to find such a route,just output “-1”.
 

Sample Input
5 8 5
1 2 2
1 5 3
1 3 4
2 4 7
2 5 6
2 3 5
3 5 1
4 5 1
2
2 3
4 3 4
1 2 3
1 3 4
2 3 2
1
1
 

Sample Output
1
-1
 

Author
dandelion
 

Source
2009浙江大学计算机研考复试(机试部分)——全真模拟
 

Recommend
lcy
原来看成是双向路径了,结果WA,看了别人的代码后才知道是单向的(也不知道哪儿说了HDU <wbr>2680 <wbr> <wbr>Choose <wbr>the <wbr>best <wbr>route),Dijkstra算法,注意一点就是把和初始点看为零,然后与他相近的点距离他均为0,也不知道是不是我的代码错了,没改之前怎么交都是TLE,改后就对了
代码:
#include<stdio.h>
int map[1008][1008],N,flag[1008],time[1008],target;
int Dijkstra(int V)
{
int i,j,min,k;
for(i=0;i<=N;i++)
{
flag[i]=0;
time[i]=map[V][i];
}
flag[V]=1;
time[V]=0;
for(i=0;i<=N;i++)
{
min=100000000;
for(j=0;j<=N;j++)
if(flag[j]==0&&min>time[j])
{
min=time[j];
k=j;
}
if(min==100000000)break;
flag[k]=1;
for(j=0;j<=N;j++)
if(flag[j]==0&&time[j]>time[k]+map[k][j])
time[j]=time[k]+map[k][j];
}
return time[target];
}
int main()
{
int i,m,j,a,b,c,num;
while(scanf("%d%d%d",&N,&m,&target)!=EOF)
{
for(i=0;i<=N;i++)
for(j=0;j<=N;j++)
map[i][j]=100000000;
for(i=1;i<=m;i++)
{
scanf("%d%d%d",&a,&b,&c);
if(map[a][b]>c)
map[a][b]=c;
// if(map[b][a]>c)
// map[b][a]=c;
}
scanf("%d",&num);
for(i=0;i<num;i++)
{
scanf("%d",&a);
// map[a][0]=
map[0][a]=0;
}
b=Dijkstra(0);
if(b!=100000000)
printf("%d\n",b);
else printf("-1\n");
}
return 0;
}
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