結果
| 問題 |
No.114 遠い未来
|
| ユーザー |
anta
|
| 提出日時 | 2014-12-29 01:42:11 |
| 言語 | C++11(廃止可能性あり) (gcc 13.3.0) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 3,434 bytes |
| コンパイル時間 | 986 ms |
| コンパイル使用メモリ | 95,320 KB |
| 実行使用メモリ | 6,944 KB |
| 最終ジャッジ日時 | 2024-06-13 00:44:25 |
| 合計ジャッジ時間 | 1,835 ms |
|
ジャッジサーバーID (参考情報) |
judge4 / judge1 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | AC * 21 WA * 4 |
コンパイルメッセージ
main.cpp: In function ‘int main()’:
main.cpp:48:14: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
48 | scanf("%d%d%d", &N, &M, &T);
| ~~~~~^~~~~~~~~~~~~~~~~~~~~~
main.cpp:54:22: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
54 | scanf("%d%d%d", &a, &b, &c), -- a, -- b;
| ~~~~~^~~~~~~~~~~~~~~~~~~~~~
main.cpp:64:22: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
64 | scanf("%d", &v), -- v;
| ~~~~~^~~~~~~~~~
ソースコード
#define _CRT_SECURE_NO_WARNINGS
#include <string>
#include <vector>
#include <algorithm>
#include <numeric>
#include <set>
#include <map>
#include <queue>
#include <iostream>
#include <sstream>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <cctype>
#include <cassert>
#include <limits>
#include <functional>
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i))
#define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i))
#if defined(_MSC_VER) || __cplusplus > 199711L
#define aut(r,v) auto r = (v)
#else
#define aut(r,v) typeof(v) r = (v)
#endif
#define each(it,o) for(aut(it, (o).begin()); it != (o).end(); ++ it)
#define all(o) (o).begin(), (o).end()
#define pb(x) push_back(x)
#define mp(x,y) make_pair((x),(y))
#define mset(m,v) memset(m,v,sizeof(m))
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3fLL
using namespace std;
typedef vector<int> vi; typedef pair<int,int> pii; typedef vector<pair<int,int> > vpii;
typedef long long ll; typedef vector<long long> vl; typedef pair<long long,long long> pll; typedef vector<pair<long long,long long> > vpll;
typedef vector<string> vs; typedef long double ld;
template<typename T, typename U> inline void amin(T &x, U y) { if(y < x) x = y; }
template<typename T, typename U> inline void amax(T &x, U y) { if(x < y) x = y; }
inline int nextCombination(int comb) {
int x = comb & -comb, y = comb + x;
return (((comb & ~y) / x) >> 1) | y;
}
int main() {
int N, M, T;
scanf("%d%d%d", &N, &M, &T);
vector<vector<int> > g(N, vector<int>(N, INF));
rep(i, N)
g[i][i] = 0;
rep(i, M) {
int a, b, c;
scanf("%d%d%d", &a, &b, &c), -- a, -- b;
amin(g[a][b], c);
amin(g[b][a], c);
}
rep(k, N) rep(i, N) rep(j, N)
amin(g[i][j], g[i][k] + g[k][j]);
vi required(T);
vector<bool> important(N, false);
rep(i, T) {
int v;
scanf("%d", &v), -- v;
required[i] = v;
important[v] = true;
}
if(T == 1) {
puts("0");
return 0;
}
vi steiners;
rep(i, N) if(!important[i])
steiners.push_back(i);
int X = steiners.size();
int ans = INF;
if(X <= 30) {
vi subsets;
subsets.push_back(0);
rer(p, 1, min(X, 3)) for(int i = (1 << p)-1; i < (1 << X); i = nextCombination(i))
subsets.push_back(i);
vector<int> vertices;
rep(usei, subsets.size()) {
int use = subsets[usei];
vertices = required;
rep(i, X) if(use >> i & 1)
vertices.push_back(steiners[i]);
int V = vertices.size();
vector<bool> vis(V, false);
vector<int> cost(V, INF);
int totalcost = 0;
cost[0] = 0;
rep(k, V) {
pii p(INF, -1);
rep(i, V) if(!vis[i])
amin(p, mp(cost[i], i));
int i = p.second, v = vertices[i];
vis[i] = true;
totalcost += cost[i];
if(totalcost >= ans) break;
rep(j, V)
amin(cost[j], g[v][vertices[j]]);
}
amin(ans, totalcost);
}
}else {
vector<int> dp((1 << T) * N, INF);
rep(p, T) rep(q, N)
dp[(1 << p) * N + q] = g[required[p]][q];
rep(S, 1 << T) {
if(!(S & (S-1))) continue;
rep(p, N) {
int x = INF;
for(int E = (S-1) & S; E > 0; (-- E) &= S)
amin(x, dp[E * N + p] + dp[(S - E) * N + p]);
dp[S * N + p] = x;
}
rep(p, N) {
int x = INF;
rep(q, N)
amin(x, dp[S * N + q] + g[p][q]);
dp[S * N + p] = x;
}
}
int U = (1 << T) - 1;
rep(S, 1 << T) rep(q, N)
amin(ans, dp[S * N + q] + dp[(U - S) * N + q]);
}
printf("%d\n", ans);
return 0;
}
anta