Gym - 102141D 通项公式 最短路

题目很长,但是意思就是给你n,A,B,C,D

n表示有n个城市 A是飞机的重量 B是一个常数表示转机代价 C是单位燃油的价格 D是一个常数

假设一个点到另外一个点的距离为整数L 起飞前的油量为f  则在这途中每飞行一单位距离 就花费(f+A)/D的燃油

#include <bits/stdc++.h>
typedef long long ll;
typedef long double lb;
using namespace std;
const int MAXN=205,MAXM=80005;
int to[MAXM << 1], nxt[MAXM << 1], Head[MAXN], ed = 1;
lb cost[MAXM << 1];
int n;
inline void addedge(int u, int v, lb c) {
        to[++ed] = v;
        nxt[ed] = Head[u];
        Head[u] = ed;
        cost[ed] = c;
}
struct HeapNode {
        int u;
        lb d;
        bool operator < (const HeapNode& rhs) const {
                return d > rhs.d;
        }
} zz;
lb mindist[MAXN];
bool vis[MAXN];
priority_queue<HeapNode> que;
void Hijkstra(int s) {
        for (int i = 0; i <= n; i++) {
                mindist[i] = 4000000000000000000.0;
        }
        mindist[s] = 0.0;
        memset(vis, 0, sizeof(vis));
        zz.d = 0, zz.u = s;
        que.push(zz);
        while (!que.empty()) {
                HeapNode x = que.top();
                que.pop();
                int u = x.u;
                if (vis[u] || mindist[u] != x.d) {
                        continue;
                }
                vis[u] = true;
                for (int v, i = Head[u]; i; i = nxt[i]) {
                        v = to[i];
                        if (mindist[v] > mindist[u] + cost[i]) {
                                mindist[v] = mindist[u] + cost[i];
                                //p[v]=u;
                                zz.d = mindist[v], zz.u = v;
                                que.push(zz);
                        }
                }
        }
}
void init(int x) {
        ed = 1;
        for (int i = 1; i <= x; i++) {
                Head[i] = 0;
        }
}
struct node {
        ll x, y;
} P[MAXN];
int main() {
        int TC;
        scanf("%d", &TC);
        while (TC--) {
                ll A, B, C, D;
                scanf("%d %lld %lld %lld %lld", &n, &A, &B, &C, &D);
                init(n);
                for (int i = 1; i <= n; i++) {
                        scanf("%lld %lld", &P[i].x, &P[i].y);
                }
                for (int i = 1; i <= n; i++) {
                        for (int j = i + 1; j <= n; j++) {
                                ll ax = abs(P[i].x - P[j].x);
                                ll ay = abs(P[i].y - P[j].y);
                                ll dis = ax * ax + ay * ay;
                                dis = ceil(sqrt(dis));
                                lb c = (pow((lb)D / (D - 1.0), dis) - 1.0) * A * C + B;
                                addedge(i,j,c),addedge(j,i,c);
                        }
                }
                Hijkstra(1);
                printf("%.10Lf\n", mindist[n]);
        }
        return 0;
}

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