HDU-2680-Choose the best route(クラシック最短問題dijkstraアルゴリズム!!)
Choose the best route
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7671 Accepted Submission(s): 2524
Problem Description
One day , Kiki wants to visit one of her friends. As she is liable to carsickness , she wants to arrive at her friend’s home as soon 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 find out the least time Kiki needs to spend. To make it easy, if the city have n bus stations ,the stations will been expressed as an integer 1,2,3…n.
Input
There are several test cases.
Each case begins with three integers n, m and s,(n<1000,m<20000,1=Then follow m lines ,each line contains three integers p , q , t (0Then a line with an integer w(0
Output
The output contains one line for each data set : the least time Kiki needs to spend ,if it’s impossible to find such a route ,just output “-1”.
Sample Input
Sample Output
Author
dandelion
Source
2009浙江大学のコンピュータは試験して再び試験します
ACコード:
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 7671 Accepted Submission(s): 2524
Problem Description
One day , Kiki wants to visit one of her friends. As she is liable to carsickness , she wants to arrive at her friend’s home as soon 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 find out the least time Kiki needs to spend. To make it easy, if the city have n bus stations ,the stations will been expressed as an integer 1,2,3…n.
Input
There are several test cases.
Each case begins with three integers n, m and s,(n<1000,m<20000,1=
Output
The output contains one line for each data set : the least time 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浙江大学のコンピュータは試験して再び試験します
ACコード:
#include <cstdio>
#include <cstring>
#include <algorithm>
#define INF 0x7f7f7f7f
using namespace std;
const int MAX = 1005;
int n, m, s;
int map[MAX][MAX], dis[MAX], vis[MAX];
void dijk(int s)
{
for(int i=1; i<=n; i++) dis[i] = map[s][i];
vis[s] = 1; dis[s] = 0;
for(int i=1; i<n; i++)
{
int min = INF, pos;
for(int j = 1; j <= n; j++)
if(!vis[j] && dis[j] < min)
min = dis[pos = j];
if(min == INF) break;
vis[pos] = 1;
for(int j=1; j<=n; j++)
if(!vis[j] && dis[pos] + map[pos][j] < dis[j])
dis[j] = dis[pos] + map[pos][j];
}
}
int main()
{
while(scanf("%d %d %d", &n, &m, &s) != EOF)
{
memset(vis, 0, sizeof(vis));
memset(map, 0x7f, sizeof(map));
while(m--)
{
int x, y, w;
scanf("%d %d %d", &x, &y, &w);
if(w < map[y][x]) map[y][x] = w; // ，
}
dijk(s);
int a, b, ans = INF;
scanf("%d", &b);
for(int i=1; i<=b; i++)
{
scanf("%d", &a);
ans = min(ans, dis[a]);
}
if(ans != INF)printf("%d
", ans);
else printf("-1
");
}
return 0;
}