Hdoj--2680--choose the best route

Choose the best routeTime limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 10372 Accepted Submission (s): 3342


Problem Descriptionone 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 your a map of the city ' s traffic route, and the stations which is near Kiki's home so, she can take. Suppose Kiki can change the bus at any station. Please find the least time Kiki needs to spend. To make it easy, if the city has n bus stations, the stations would been expressed as an integer 1,2,3...N.
Inputthere is several test cases.
Each case begins with three integers n, m and S, (N<1000,M<20000,1=<S<=N) n stands for the number of bus statio NS in this city and M stands for the number of directed ways between bus stations. (maybe there is several ways between, bus stations.) s stands for the bus station 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 p to station Q there is a and it'll costs T minutes.
Then a line with an integer w (0<w<n), means the number of stations Kiki can take at the beginning. Then follows W integers stands for these stations.

Outputthe 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 51 2 21 5 31 3 42 4 72 5 62 3 53 5 14 5 122 34 3 41 2 31 3 42 3 211

Sample Output

1-1

Test Instructions: Kiki wants to visit her friends, but the goods are easy to get carsick, so find a route that takes the least time. Now give you the roadmap to find a route that takes the least amount of time from the starting point near her home (there can be multiple) to the terminal near the friend's home (only one). Each site is numbered from 1 to N.  idea: At first I was directly without the brain using the DIJ algorithm to do the results in the slightest surprise (TLE), fuck. Look at other people's thinking to build should end when the beginning of the reverse composition, note that this is a one-way map. AC Code:

#include <stdio.h> #include <string.h> #define INF 0x3f3f3f3f#include<algorithm>using namespace std; int vis[1010],map[1010][1010],dis[1010],n,ans[1010],num,beg;void init () {for (Int. i=1;i<=n;i++) for (int j=0;j<= n;j++) {if (i==j) Map[i][j]=map[j][i]=0;elsemap[i][j]=map[j][i]=inf;}} void Dijkstra () {int K=0,flag=0,i;memset (vis,0,sizeof (VIS)); for (i=1;i<=n;i++) dis[i]=map[beg][i];vis[beg]=1;for (i=1;i<=n;i++) {int j,key,temp=inf;for (int j=1;j<=n;j++) if (!vis[j]&&temp>dis[j]) temp=dis[key=j];if (Temp==INF) { break;} vis[key]=1;for (int j=1;j<=n;j++) if (!vis[j]&&dis[j]>dis[key]+map[key][j]) dis[j]=dis[key]+map[key][ j];}} int main () {int m;while (scanf ("%d%d%d", &n,&m,&beg)!=eof) {int I;memset (ans,inf,sizeof (INF)); init ();// Note to initialize. for (i=1;i<=m;i++) {int a,b,cost;scanf ("%d%d%d", &a,&b,&cost), if (map[b][a]>cost)//filter out the same edge. Reverse framing. Map[b][a]=cost;} scanf ("%d", &num);d Ijkstra ();//Find the location of each end point directly. for (i=0;i<num;i++) {int end;scanf ("%d", &end);Ans[i]=dis[end];} Sort (ans,ans+num), if (Ans[0]==inf)//To determine if there is such a value, does not exist if the ANS array is sure that each is an INF. printf (" -1\n"); elseprintf ("%d\n", Ans[0]);} return 0;}

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Hdoj--2680--choose the best route