Lazy Running

思路

还是利用同余的思想，假设存在一条长度为$k$的路，那么也一定存在一条$k+base$的路$base=2\ast min(d1,d2)$。

$dis[i][j]=x$表示的是，从$2->i$点$x\equiv j(\mathrm{m}\mathrm{o}\mathrm{d}base)$的满足条件的最小的$x$，所以我们只要求出所有的$dis[2][i]$，再通过同余的性质去得到我们的最短路的花费，对于$dis[2][i]>k$我们取$min(ans,dis[2][i])$，否则的话，我们取$min(ans,dis[2][i]+(k-dis[2][i]+mid-1)/mod\ast mod)$，之后我们就可以得到我们的正确解了

代码

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define mp make_pair
#define pb push_back
#define endl '\n'using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}void print(ll x) {if(x < 10) {putchar(x + 48);return ;}print(x / 10);putchar(x % 10 + 48);
}const int N = 6e4 + 10;ll dis[5][N], k, d1, d2, d3, d4, mod;vector< pair< int, ll > > G[5];void Dijkstra() {priority_queue< pair< ll, int >, vector< pair< ll, int > >, greater< pair< ll, int > > > q;q.push(mp(0, 2));memset(dis, 0x3f, sizeof dis);dis[2][0] = 0;while(q.size()) {auto temp = q.top();q.pop();if(temp.first > dis[temp.second][temp.first % mod]) continue;for(auto i : G[temp.second]) {int to = i.first;ll w = i.second + temp.first;if(dis[to][w % mod] > w) {dis[to][w % mod] = w;q.push(mp(w, to));}}}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int t = read();while(t--) {k = read(), d1 = read(), d2 = read(), d3 = read(), d4 = read();mod = min(d1, d2) * 2;for(int i = 1; i <= 4; i++) G[i].clear();G[1].pb(mp(2, d1)), G[1].pb(mp(4, d4));G[2].pb(mp(3, d2)), G[2].pb(mp(1, d1));G[3].pb(mp(2, d2)), G[3].pb(mp(4, d3));G[4].pb(mp(3, d3)), G[4].pb(mp(1, d4));Dijkstra();ll ans = 0x3f3f3f3f3f3f3f3f;for(int i = 0; i < mod; i++) {if(k < dis[2][i]) ans = min(ans, dis[2][i]);else {ans = min(ans, dis[2][i] + (k - dis[2][i] + mod - 1) / mod * mod);}}printf("%lld\n", ans);}return 0;
}