이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#define x first
#define y second
#define pb push_back
#define mp make_pair
#define all(v) v.begin(),v.end()
#define sz(v) (int)v.size()
#define up_b upper_bound
#define low_b lower_bound
#define nl '\n'
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>ordered_set;
template<class T,int SZ> struct BIT{
T t[SZ];
void upd(int x,T y){
for(int i=x;i<SZ;i=(i|(i+1))){
t[i]+=y;
}
}
T pref(int x){
T res=0;
for(int i=x;i>=0;i=(i&(i+1))-1){
res+=t[i];
}
return res;
}
T get(int l,int r){
return pref(r)-pref(l-1);
}
};
template<class T,int SZ> struct ST{
T t[4*SZ];
void build(T a[],int v,int tl,int tr){
if(tl==tr){
t[v]=a[tl];
return ;
}
int tm=(tl+tr)/2;
build(a,v*2,tl,tm);
build(a,v*2+1,tm+1,tr);
t[v]=t[v*2]+t[v*2+1];
}
void upd(int v,int tl,int tr,int pos,T val){
if(tl==tr){
t[v]+=val;
return ;
}
int tm=(tl+tr)/2;
if(pos<=tm)upd(v*2,tl,tm,pos,val);
else upd(v*2+1,tm+1,tr,pos,val);
t[v]=t[v*2]+t[v*2+1];
}
T get(int v,int tl,int tr,int l,int r){
if(tl>r||l>tr)return 0;
if(l<=tl&&tr<=r)return t[v];
int tm=(tl+tr)/2;
return get(v*2,tl,tm,l,r)+get(v*2+1,tm+1,tr,l,r);
}
};
const int N=111;
const int K=1e3+11;
const int W=1e3+11;
const int inf=INT_MAX;
const ll INF=1e18;
const ll mod=1e9+7;
const ld EPS=1e-9;
const int dx[4]={0,0,1,-1};
const int dy[4]={1,-1,0,0};
vector<pair<int,int>>g[N];
int b[N][K],s[N][K];
ld d[N];
int main(){
ios_base::sync_with_stdio(0);
cin.tie(0),cout.tie(0);
int n,m,k;
cin>>n>>m>>k;
for(int i=1;i<=n;i++){
for(int j=1;j<=k;j++){
cin>>b[i][j]>>s[i][j];
}
}
for(int i=1;i<=m;i++){
int u,v,w;
cin>>u>>v>>w;
g[u].pb(mp(v,w));
g[v].pb(mp(u,w));
}
set<pair<int,int>>q;
q.insert(mp(0,1));
d[1]=0;
while(!q.empty()){
pair<int,int>p=*q.begin();
q.erase(p);
int v=p.y;
for(int i=0;i<sz(g[v]);i++){
int to=g[v][i].x;
if(d[to]>d[v]+g[v][i].y){
q.erase(mp(d[to],to));
d[to]=d[v]+g[v][i].y;
q.insert(mp(d[to],to));
}
}
}
ld ans=0;
for(int i=2;i<=n;i++){
for(int j=1;j<=k;j++){
ld cost=s[i][j]-b[1][j];
if(cost<=0)continue;
ans=max(ans,cost/(2*d[i]));
}
}
cout<<ans;
}
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