2020 cattle off winter algorithm base Camp 3 J- beef Po can dream Go (DP + Floyd)

Links: https://ac.nowcoder.com/acm/contest/3004/J

Ideas:

Pretreatment with the first flyod dis [i] [j] represents the shortest distance from point i to j

After starting DP, dp [i] denote the i-th wizard may appear harvest the maximum combat effectiveness

Then dp [i] to obtain the transfer equation dp [i] = max (dp  [i], dp [ij] + a [i] .val) represents Po may dream before i j number of occurrences of

Then this will be the time complexity is K ² , but we can see that the distance between any two points on the map are less than 200, it can only be a maximum of 200 j, so it reduced complexity O (200 * K )

Note that the time difference is determined also dreams can occur two Po is greater than the distance between two points is equal to

#include<iostream>
#include<algorithm>
#include<cstring>
 using namespace std;
 typedef long long ll;
 const int maxn=1e5+10;
 const int inf=0x3f3f3f3f;
 int dis[205][205],n,m,u,v,k;
 ll ans=0,dp[maxn];
 struct pok{
     int t,pos,v;
 }a[maxn];
 int cmp(pok a,pok b){return a.t<b.t;}
 void floyd()
 {
      for(int k=1;k<=n;++k)
        for(int i=1;i<=n;++i)
            for(int j=1;j<=n;++j)
                dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);
 }
 int main()
 {
     scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++) 
            dis[i][j]= i==j?0:inf;
     for(int i=1;i<=m;i++){
         scanf("%d%d",&u,&v);
         dis[u][v]=dis[v][u]=1;
     }
    floyd();
    scanf("%d",&k);
    for(int i=1;i<=k;i++)
        scanf("%d%d%d",&a[i].t,&a[i].pos,&a[i].v);
    sort(a+1,a+1+k,cmp);
    a[0].t=0,a[0].pos=1,a[0].v=0;
    for(int i=1;i<=k;i++){
        dp[i]=-inf;
        for(int j=1;j<=200&&i>=j;j++){
            if(a[i].t-a[i-j].t>=dis[a[i].pos][a[i-j].pos])
                dp[i]=max(dp[i],dp[i-j]+a[i].v);
        }
        ans=max(dp[i],ans);
    }
    cout<<ans<<endl;
    return 0;
}