CCPC Harbin 2021 G, Damaged Bicycle

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whiteawll 2022-08-16 14:45:36 博主文章分类：练习题目 ©著作权

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#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;   
typedef vector<ll> Vll;               
typedef vector<pair<int, int> > vpii;
typedef vector<pair<ll, ll> > vpll;                        

const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const double pi  = acos(-1.0);

inline ll qmi (ll a, ll b) {
    ll ans = 1;
    while (b) {
        if (b & 1) ans = ans * a%mod;
        a = a * a %mod;
        b >>= 1;
    }
    return ans;
}
inline int read () {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f?-x:x;
}
template<typename T> void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x/10);
    putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
    return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
    return (a + b) %mod;
}
inline ll inv (ll a) {
    return qmi(a, mod - 2);
}

const int N = 1e6 + 10;

struct edge {
    int v, w;
};
struct node {
    int u, dis;
};
bool operator< (const node & q, const node & p) {
    return q.dis > p.dis;
}
vector<edge> g[N];
int n, m, k;
double p[N], t, r;
int a[N];
int vis[26][N], dist[26][N];

void dijk (int id, int x) {
     memset(dist[id], 0x3f, sizeof dist[id]);
     dist[id][x] = 0;
     priority_queue<node> heap;
     heap.push({x, 0});
     while (heap.size()) {
         auto t = heap.top(); heap.pop();
         int u = t.u, dis = t.dis;
         if (vis[id][u]) continue;
         vis[id][u] = 1;
         dist[id][u] = dis;
         for (auto t : g[u]) {
             int v = t.v, w = t.w;
             if (dist[id][v] > w + dis) {
                 dist[id][v] = w + dis;
                 heap.push({v, dist[id][v]});
             }
         }
     }
}
double dp[26][N];
double dfs (int u, int state) {
    if (dp[u][state])  return dp[u][state];
    double res = 1.0 * p[u] * dist[u][n] / t + (1 - p[u]) * dist[u][n] / r;
    for (int i = 1; i <= k; i ++) {
        if (state & (1 << (i - 1))) continue;
        res = min (res, 1.0 * (1 - p[u]) *dist[u][n] / r + 
         p[u] * (dist[u][a[i]] / t + dfs(i, state | (1 << (i - 1)))));
    }
    return dp[u][state] = res;
}

void solve() {
    cin >>t >>  r >> n >> m;
    for (int i = 1; i <= m; i ++) {
        int u, v, w;
        scanf("%d%d%d", & u, & v, & w);
        g[u].pb({v, w});
        g[v].pb({u, w});
    }
    cin >> k;
    for (int i = 1; i<= k; i ++) {
        cin >> a[i] >> p[i];
        p[i] /= 100;
        dijk(i, a[i]);
    }
    
    dijk(19, 1);
    if (dist[19][n] == 0x3f3f3f3f) {
        cout <<"-1\n";
        return;
    }
    p[19] = 1;
    printf("%.6lf\n", dfs(19, 0));
}
int main () {
    // ios::sync_with_stdio(0),cin.tie(0), cout.tie(0);
    int t;
    t =1;
    //cin >> t;
    while (t --) solve();
    return 0;
}