HDU2145 zz's Mysterious Present 解题报告 ——最短路dijkstra算法

zz's Mysterious Present

Time Limit: 10000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 854    Accepted Submission(s): 171


Problem Description

There are m people in n cities, and they all want to attend the party which hold by zz. They set out at the same time, and they all will choose the best way they think, but due to someone take a ride, someone drive, and someone take a taxi, they have different
speed. Can you find out who will get zz's mysterious present? The first one get the party will get the present . If there are several people get at the same time, the one who stay in the city which is farther from the city where is zz at begin will get the
present. If there are several people get at the same time and the distance from the city he is at begin to the city where zz is, the one who has the larger number will get the present.

 

Input

The first line: three integers n, m and k. m is the total number of the people, and n is the total number of cities, and k is the number of the way.(0<n<=300, 0<m<=n, 0<k<5000)

The second line to the (k+1)th line: three integers a, b and c. There is a way from a to b, and the length of the way is c.(0<a,b<=n, 0<c<=100)

The (k+2)th line: one integer p(0<p<=n), p is the city where zz is.

The (k+3)th line: m integers. the ith people is at the place p[i] at begin.(0<p[i]<=n)

The (k+4)th line: m integers. the speed of the ith people is speed[i];(0<speed[i]<=100) 
All the ways are directed.

 

Output

For each case, output the one who get the present in one line. If no one can get the present, output "No one".

 

Sample Input

3 1 3
1 2 2
1 3 3
2 3 1
3
2
1

 

Sample Output

1

 

这道题在最短路的基础上加入了速度和时间，可以直接将终点P看为起点，便可以用Dijkstra解题。
因为题目说明所以路是有向的，所以输入时应注意顺序。

#include<stdio.h>
#include<string.h>

const long inf =999999999;

int map[301][301];
double	time[301];
int speed[301];
int pla[301];
int p;
long dis[301];

void Dijkstra(int p,int n){
bool used[301]={false};
long min,next;
memset(dis,-1,sizeof(dis));
dis[p]=0;

for(long i=1;i<=n;i++)
{
min=inf;
for(long j=1;j<=n;j++)
if(!used[j] && dis[j]!=-1 && dis[j]<min)
{
min=dis[j];
next=j;
}
if(min!=inf)
{
used[next]=true;
for(long j=1;j<=n;j++)
if(!used[j] && map[next][j]!=-1 && (dis[j]>map[next][j]+dis[next] || dis[j]==-1))
dis[j]=map[next][j]+dis[next];
}

}
}

int main(void){
int n,m,k;//city,peo,way
int o,p,q;//xunhuan
int a,b,c;//input
while(scanf("%d %d %d",&n,&m,&k)!=EOF){
for(o=1;o<=n;++o)
for(p=1;p<=n;++p)
map[o][p]=inf;
for(o=1;o<=k;o++){
scanf("%d %d %d",&a,&b,&c);
if(map[b][a]>c){
map[b][a]=c;
}
}

scanf("%d",&p);
Dijkstra(p,n);

long min=1;
for(o=1;o<=m;++o)
scanf("%d",&pla[o]);
for(o=1;o<=m;++o)
scanf("%d",&speed[o]);
for(o=1;o<=m;++o){
time[o]=((double)(dis[pla[o]])/(double)(speed[o]));
if(time[min]>time[o])
min=o;
else if(time[min]==time[o]){
if(dis[pla[min]]<=dis[pla[o]])
min=o;
}
}
if(dis[pla[min]]==inf)
printf("No one\n");
else
printf("%d\n",min);
}
return 0;
}