結果
| 問題 |
No.1 道のショートカット
|
| コンテスト | |
| ユーザー |
 shimomire
|
| 提出日時 | 2015-01-02 20:58:50 |
| 言語 | C++11（廃止可能性あり） (gcc 13.3.0) |
| 結果 |
RE
|
| 実行時間 | - |
| コード長 | 8,291 bytes |
| コンパイル時間 | 1,201 ms |
| コンパイル使用メモリ | 110,092 KB |
| 実行使用メモリ | 19,964 KB |
| 最終ジャッジ日時 | 2024-07-08 03:40:04 |
| 合計ジャッジ時間 | 6,969 ms |
|
ジャッジサーバーID （参考情報） |
judge2 / judge1 |
（要ログイン）
| ファイルパターン | 結果 |
|---|---|
| sample | RE * 4 |
| other | AC * 12 WA * 5 RE * 23 |
ソースコード
#include <cassert>// c
#include <iostream>// io
#include <iomanip>
#include <fstream>
#include <sstream>
#include <vector>// container
#include <map>
#include <set>
#include <queue>
#include <bitset>
#include <stack>
#include <algorithm>// other
#include <complex>
#include <numeric>
#include <functional>
//#include <random>
//#include <regex>
using namespace std;
//using i32=int32_t;using i64=int64_t;using ll =i64;using uint=uint32_t;using ull=uint64_t;
//template<typename T> using matrix=vector<vector<T> >;
typedef long long ll;
#define ALL(c) (begin(c)),(end(c))
#define REP(i,n) FOR(i,0,n)
#define REPr(i,n) FORr(i,0,n)
#define FOR(i,l,r) for(int i=(int)(l);i<(int)(r);++i)
#define FORr(i,l,r) for(int i=(int)(r)-1;i>=(int)(l);--i)
#define EACH(it,o) for(auto it = (o).begin(); it != (o).end(); ++it)
#define IN(l,v,r) ((l)<=(v) && (v)<(r))
#define UNIQUE(v) v.erase(unique(ALL(v)),v.end())
//debug
#define DUMP(x) cerr << #x << " = " << (x)
#define LINE() cerr<< " (L" << __LINE__ << ")"
class range {
private:
struct Iter{
int v;
int operator*(){return v;}
bool operator!=(Iter& itr) {return v < itr.v;}
void operator++() {++v;}
};
Iter i, n;
public:
range(int n) : i({0}), n({n}) {}
range(int i, int n) : i({i}), n({n}) {}
Iter& begin() {return i;}
Iter& end() {return n;}
};
//output
//template<typename T> ostream& operator << (ostream& os, const vector<T>& as){REP(i,as.size()){if(i!=0)os<<" "; os<<as[i];}return os;}
//template<typename T> ostream& operator << (ostream& os, const vector<vector<T>>& as){REP(i,as.size()){if(i!=0)os<<endl; os<<as[i];}return os;}
//template<typename T> ostream& operator << (ostream& os, const set<T>& ss){for(auto a:ss){if(a!=ss.begin())os<<" "; os<<a;}return os;}
//template<typename T1,typename T2> ostream& operator << (ostream& os, const pair<T1,T2>& p){os<<p.first<<" "<<p.second;return os;}
//template<typename K,typename V> ostream& operator << (ostream& os, const map<K,V>& m){bool isF=true;for(auto& p:m){if(!isF)os<<endl;os<<p;isF=false;}return os;}
//template<typename T1> ostream& operator << (ostream& os, const tuple<T1>& t){os << get<0>(t);return os;}
//template<typename T1,typename T2> ostream& operator << (ostream& os, const tuple<T1,T2>& t){os << get<0>(t)<<" "<<get<1>(t);return os;}
//template<typename T1,typename T2,typename T3> ostream& operator << (ostream& os, const tuple<T1,T2,T3>& t){os << get<0>(t)<<" "<<get<1>(t)<<" "<<get<2>(t);return os;}
//template<typename T1,typename T2,typename T3,typename T4> ostream& operator << (ostream& os, const tuple<T1,T2,T3,T4>& t){os << get<0>(t)<<" "<<get<1>(t)<<" "<<get<2>(t)<<" "<<get<3>(t);return os;}
//template<typename T1,typename T2,typename T3,typename T4,typename T5> ostream& operator << (ostream& os, const tuple<T1,T2,T3,T4,T5>& t){os << get<0>(t)<<" "<<get<1>(t)<<" "<<get<2>(t)<<" "<<get<3>(t)<<" "<<get<4>(t);return os;}
//template<typename T1,typename T2,typename T3,typename T4,typename T5,typename T6> ostream& operator << (ostream& os, const tuple<T1,T2,T3,T4,T5,T6>& t){os << get<0>(t)<<" "<<get<1>(t)<<" "<<get<2>(t)<<" "<<get<3>(t)<<" "<<get<4>(t)<<" "<<get<5>(t);return os;}
//template<typename T1,typename T2,typename T3,typename T4,typename T5,typename T6,typename T7> ostream& operator << (ostream& os, const tuple<T1,T2,T3,T4,T5,T6,T7>& t){os << get<0>(t)<<" "<<get<1>(t)<<" "<<get<2>(t)<<" "<<get<3>(t)<<" "<<get<4>(t)<<" "<<get<5>(t)<<" "<<get<6>(t);return os;}
//input
char tmp[1000];
#define nextInt(n) scanf("%d",&n)
#define nextLong(n) scanf("%lld",&n) //I64d
#define nextDouble(n) scanf("%lf",&n)
#define nextChar(n) scanf("%c",&n)
#define nextString(n) scanf("%s",tmp);n=tmp
// values
template<typename T> T INF(){assert(false);};
template<> int INF<int>(){return 1<<28;};
template<> ll INF<ll>(){return 1LL<<58;};
template<> double INF<double>(){return 1e16;};
template<class T> T EPS(){assert(false);};
template<> int EPS<int>(){return 1;};
template<> ll EPS<ll>(){return 1LL;};
template<> double EPS<double>(){return 1e-8;};
template<> long double EPS<long double>(){return 1e-8;};
template<typename T,typename U> T pmod(T v,U M){return (v%M+M)%M;}
ll gcd_positive(ll a,ll b) { return b == 0 ? a : gcd_positive(b,a%b); }
ll gcd(ll a,ll b) { return gcd_positive(abs(a), abs(b)); }
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
namespace EGraph{
typedef ll Cost;Cost CINF=INF<Cost>();
struct Edge{
int from,to;Cost cost;int money;
Edge(int from,int to,Cost cost,int money)
: from(from),to(to),cost(cost),money(money){};
bool operator<(Edge r) const{ return cost<r.cost;}
bool operator>(Edge r) const{ return cost>r.cost;}
};
typedef vector<vector<Edge> > Graph;
}
using namespace EGraph;
namespace ShortestPath{
using namespace EGraph;
struct Task{
int prev,pos;Cost cost;
Task(int prev,int pos,Cost cost)
:prev(prev),pos(pos),cost(cost){};
bool operator>(const Task& r) const{ return cost > r.cost;}
};
// 最大も可 負の閉路が判定可能
//O(V*E)
vector<Cost> bellmanFord(const Graph& g,const int s,vector<int>& prev){
const int V=g.size();
vector<Cost> d(V,CINF);d[s] = 0;
fill(ALL(prev),-2);
REP(loop,V){
REP(v,V)EACH(e,g[v])if(d[v] != CINF)if(d[e->to] > d[v] + e->cost){
d[e->to] = d[v] + e->cost;
prev[e->to] = e->from;
if(loop==V-1)d[e->to] = -CINF;
}
}
return d;
}
vector<Cost> bellmanFord(const Graph& g,const int s){
vector<int> prev(g.size());return bellmanFord(g,s,prev);
}
//verified by codoforces 144D http://codeforces.com/contest/144/submission/4976825
// // 負の辺がない
// // O(E*logV)
vector<Cost> dijkstra(const Graph& g,const int s,vector<int>& prev){
const int V=g.size();
vector<Cost> d(V,CINF);d[s]=0;
fill(ALL(prev), -2);
priority_queue<Task,vector<Task>,greater<Task> > que;que.push(Task(-1,s,0));// [ ,e,,f, ] <=> e.cost < e.cost
vector<bool> visited(V);
while(!que.empty()){
Task task=que.top();que.pop();
if(visited[task.pos])continue;
visited[task.pos]=true;
prev[task.pos]=task.prev;
EACH(e,g[task.pos])if(d[e->to]>d[e->from]+e->cost){
d[e->to]=d[e->from]+e->cost;
que.push(Task(e->from,e->to,d[e->to]));
}
}
return d;
}
vector<Cost> dijkstra(const Graph& g,const int s){
vector<int> prev(g.size());return dijkstra(g,s,prev);
}
// verified by poj3255
// http://poj.org/showsource?solution_id=13511085
// k番目の最短距離
// O(k*ElogV)
vector<vector<Cost> > dijkstra_k(const Graph& g,const int s,const int k){
const int V=g.size();
vector<vector<Cost> > d(V);
priority_queue<Task,vector<Task>,greater<Task> > que;que.push(Task(-1,s,0));
while(!que.empty()){
Task task=que.top();que.pop();
if(d[task.pos].size() >= k)continue;
d[task.pos].push_back(task.cost);
EACH(e,g[task.pos])que.push(Task(e->from,e->to,task.cost + e->cost));
}
return d;
}
}
using namespace ShortestPath;
class Main{
public:
int N,C,V;
int enc(int i,int c){
return i*C+c;
}
void run(){
cin >> N >> C >> V;
Graph g(N*C);
vector<Edge> es;REP(i,V)es.push_back(Edge(0,0,0,0));
REP(i,V){
cin >> es[i].from;
es[i].from--;
}
REP(i,V){
cin >> es[i].to;
es[i].to--;
}
REP(i,V)cin >> es[i].money;
REP(i,V)cin >> es[i].cost;
REP(i,V){
REP(c,C+1)if(c-es[i].money>=0){
g[enc(es[i].from,c)].push_back(Edge(enc(es[i].from,c),enc(es[i].to,c-es[i].money),es[i].cost,0));
g[enc(es[i].to,c)].push_back(Edge(enc(es[i].to,c),enc(es[i].from,c-es[i].money),es[i].cost,0));
}
}
vector<Cost> res=dijkstra(g,enc(0,C));
Cost mv=INF<Cost>();
REP(c,C)mv=min(mv,res[enc(N-1,c)]);
if(mv==INF<Cost>()) cout << -1 << endl;
else cout << mv <<endl;
}
};
int main(){
cout <<fixed<<setprecision(20);
cin.tie(0);
ios::sync_with_stdio(false);
Main().run();
return 0;
}