HDU 6071 Lazy Running（同余最短路的应用）

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 $base = 2 * min(d1, d2)$。

$dis[i][j] = x$表示的是，从 $2 -> i$点 $x \equiv j \pmod {base}$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 $min(ans, dis[2][i] + (k - dis[2][i] + mid - 1) / mod * mod)$，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;
}