[Bzoj5511] big center playground

Referred cola 1 , Hamburg, -1 , i.e. the process of seeking an absolute value does not exceed k Shortest Path.

Then find k range is only 10 , that is the only course of legal values 21 species, and therefore run again dij or SPFA (hey hey hey) can be.

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define pi pair<int,int>
 4 #define mp make_pair
 5 #define fi first
 6 #define se second
 7 #define N 10005
 8 queue<pi >q;
 9 struct ji{
10     int nex,to,len;
11 }edge[N*20];
12 int E,t,n,m,k,x,y,z,ans,d[N][41],vis[N][41],head[N],p[N];
13 void add(int x,int y,int z){
14     edge[E].nex=head[x];
15     edge[E].to=y;
16     edge[E].len=z;
17     head[x]=E++;
18 }
19 int main(){
20     scanf("%d",&t);
21     while (t--){
22         scanf("%d%d%d",&n,&m,&k);
23         memset(d,0x3f,sizeof(d));
24         memset(head,-1,sizeof(head));
25         memset(vis,0,sizeof(vis));
26         E=0;
27         for(int i=1;i<=n;i++){
28             scanf("%d",&p[i]);
29             p[i]=p[i]*2-3;
30         }
31         for(int i=1;i<=m;i++){
32             scanf("%d%d%d",&x,&y,&z);
33             add(x,y,z);
34             add(y,x,z);
35         }
36         scanf("%d%d",&x,&y);
37         d[x][p[x]+20]=0;
38         vis[x][p[x]+20]=1;
39         q.push(mp(x,p[x]+20));
40         while (!q.empty()){
41             pi o=q.front();
42             q.pop ();
43              VIS [o.fi] [o.se] = 0 ?
44              for ( int i = head [o.fi]? I! = - 1 ? I = EDGE [i] .nex) {
 45                  x = EDGE [i] .TO?
46                  the if ((abs (o.se + p [x] - 20 ) <= k) && (d [o.fi] [o.se] + EDGE [i] .len <d [x] [o.se + p [x]])) {
 47                      d [x] [o.se + p [x]] = d [o.fi] [o.se] + EDGE [i] .len?
48                      the if (! VIS [x] [o.se + p [x]]) {
 49                          q.push (mp (x, o.se + p [x]))?
50                          VIS [x] [o.se + p [x]] = 1;
51                      }
 52                  }
 53              }
 54          }
 55          years = 0x3f3f3f3f ;
56          for ( int i = 20 k; i <= k + 20 ; i ++) ans = min (years d [y] [i]);
57          if (age == 0x3f3f3f3f ) ans = - 1 ;
58          printf ( " % d \ n " , year);
59      }
 60 }

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