HDU 6071

/*
HDU 6071 - Lazy Running [ 建模,最短路 ]  |  2017 Multi-University Training Contest 4
题意：
	四个点的环，给定相邻两点距离，问从2号点出发，2号点结束，距离大于k的最短距离
	d(i,j) < 30000， k <= 1e18
分析：
	设 r = min(d(1,2), d(2,3))
	假设距离 k 可达，则 k+2*r 也可达（来回走的形式）
	故求出所有 d[i][j] 满足 d[i][j]%2r == j 的最短距离
	然后对于每一个d[2][j] 求出 d[2][j] + x*(2r) >= k 的第一个数
*/
#include <bits/stdc++.h>
using namespace std;
#define LL long long
const LL INF = (LL)(1e18)+5;
struct qnode {
    int v; LL c;
    qnode(int v = 0, LL c = 0) : v(v), c(c) {}
    bool operator < (const qnode & r) const {
        return c > r.c;
    }
};
struct Edge {
    int v; LL c;
    Edge(int v = 0, LL c = 0) : v(v), c(c) {}
};
vector<Edge> edge[5];
bool vis[5][60005];
LL dist[5][60005];
priority_queue<qnode> que;
void Dijkstra(int r)
{
    while (!que.empty()) que.pop();
    memset(vis, 0, sizeof(vis));
    for (int i = 1; i <= 4; i++)
        for (int j = 0; j <= r; j++) dist[i][j] = INF;
    dist[2][0] = r;
    que.push(qnode(2, 0));
    while (!que.empty())
    {
        auto tmp = que.top(); que.pop();
        int u = tmp.v;
        LL w = tmp.c;
        if (vis[u][w%r]) continue;
        vis[u][w%r] = 1;
        for (const auto& e : edge[u])
        {
            LL c = w + e.c;
            if (!vis[e.v][c%r] && dist[e.v][c%r] > c)
            {
                dist[e.v][c%r] = c;
                que.push(qnode(e.v, c));
            }
        }
    }
}
void addedge(int u, int v, LL w)
{
    edge[u].push_back(Edge(v, w));
    edge[v].push_back(Edge(u, w));
}
LL k, d[5], r;
int main()
{
    int t; scanf("%d", &t);
    while (t--)
    {
        for (int i = 1; i <= 4; i++) edge[i].clear();
        scanf("%lld", &k);
        for (int i = 1; i <= 4; i++) scanf("%lld", &d[i]);
        r = 2*min(d[1], d[2]);
        addedge(1, 2, d[1]); addedge(1, 4, d[4]);
        addedge(2, 3, d[2]); addedge(3, 4, d[3]);
        Dijkstra(r);
        LL ans = k+r;
        for (int i = 0; i < r; i++)
        {
            if (dist[2][i] == INF) continue;
            if (dist[2][i] >= k)
            {
                ans = min(ans, dist[2][i]);
            }
            else
            {
                LL s = k-dist[2][i];
                LL l = ((s-1)/r+1) * r + dist[2][i];
                ans = min(ans, l);
            }
        }
        printf("%lld
", ans);
    }
}