HDU 2851 Dijkstra algorithm variants

Dijkstra algorithm Template:

int visited[i]//node I if accessed is 1, there is no 0Int dist[i]//current node J to the other nodes of the shortest length int w[i][j]//edge (i,j) of the Weight value initialization: (node 1~n) memset (v,0,sizeof (v)); Dist[j]=0;dist[i]=inf; (i>=1&&i<=n&&i!=j)    cin>>a>>b>>x;     if (w[a][b]>x)//a to B may have more than one path    w[b][a]=w[a][b]=x;w[x][y]=inf; the Edge (x, y) does not exist for (i=1;i<=n;i++) {    int x,m=inf;    for (j=1;j<=n;j++)        if (!visited[j]&&dist[j]<m)        {            m=dist[j];            x=j;        }    Visited[x]=1;for (j=1;j<=n;j++) If (!visited[j])        dist[j]=min (Dist[j],dist[x]+w[x][j])}

Description

Lode Runner is a famous game, I think you played in your childhood.


Ok, now, I simple the problem. In the game, it had N horizontal roads, the ladders always stay at right side vertically, and the ladders is extending to Wards up and down with unlimited length. If ladder near or cross a road, little Wiskey would walk on the road by climbed the ladder. Roads were covered by each of the other in the x-axis; It'll be considered can passing through. Each road has a integer W means the dangerous level.

Little Wiskey must start at the most road, and he can ' t go back, he always go ahead. Wiskey want to go some roads, so he needs to know how minimum sum of dangerous would happen. You can finish the game, yes?

Input

The first integer C represents the number of cases, and C cases followed.

Each test case is contains a single integer N roads (1<=n<=) and M destinations (1<=m<=500). The next N line contains three integers Si, Ei, Wi, meaning the road build from Si to Ei, and the Wi dangerous level (0 &L t;= Si <= Ei <=, 1 <= Wi <= 1000). And the roads sorted by Ei increasing yet. The last M line contains integers mean little wiskey ' s destinations.

Output

For each questions, output the minimum sum of dangerous. If can ' t reach the goal, please print-1.

Sample Input

Sample Output

Turn the weight of a point into the weight of the edge that leads to it

#include <iostream> #include <cstring> #include <cstdio> #include <cmath> #include <string    > #include <algorithm> #define LL long long#define inf 0x3f3f3f3fusing namespace std;struct node{int l,r; int W;}    Q[2010];bool visited[2010];int dist[2010];int wx[2010][2010];int main () {int T;    cin>>t;        while (t--) {int n,m;        cin>>n>>m;        for (int i=1;i<=n;i++) scanf ("%d%d%d", &AMP;Q[I].L,&AMP;Q[I].R,&AMP;Q[I].W);        memset (wx,inf,sizeof (WX));        for (int i=1;i<=n;i++) for (int j=1;j<=n;j++) if (i==j) wx[i][j]=wx[j][i]=0; for (int i=2;i<=n;i++) for (int j=1;j<=i-1;j++) if (Q[J].R&GT;=Q[I].L) wx[    J][I]=Q[I].W;    memset (visited,0,sizeof (visited));    memset (dist,inf,sizeof (Dist));    dist[1]=0;        for (int i=1;i<=n;i++) {int x,mx=inf; for (int j=1;j<=n;j++) if (!visited[j]&&dist[j]&LT;MX) {mx=dist[j];        X=j;        } visited[x]=1;    for (int j=1;j<=n;j++) dist[j]=min (Dist[j],dist[x]+wx[x][j]);            } while (m--) {int b;            cin>>b;            if (Dist[b]==inf) cout<<-1<<endl;        else cout<<dist[b]+q[1].w<<endl; }} return 0;}

HDU 2851 Dijkstra algorithm variants