【多校训练】hdu 6071 Lazy Running 最短路径 spfa

Problem Description

In HDU, you have to run along the campus for 24 times, or you will fail in PE. According to the rule, you must keep your speed, and your running distance should not be less than K meters.

There are 4 checkpoints in the campus, indexed as p1,p2,p3 and p4. Every time you pass a checkpoint, you should swipe your card, then the distance between this checkpoint and the last checkpoint you passed will be added to your total distance.

The system regards these 4 checkpoints as a circle. When you are at checkpoint pi, you can just run to pi−1 or pi+1(p1 is also next to p4). You can run more distance between two adjacent checkpoints, but only the distance saved at the system will be counted.





Checkpoint p2 is the nearest to the dormitory, Little Q always starts and ends running at this checkpoint. Please write a program to help Little Q find the shortest path whose total distance is not less than K.

 

Input

The first line of the input contains an integer T(1≤T≤15), denoting the number of test cases.

In each test case, there are 5 integers K,d1,2,d2,3,d3,4,d4,1(1≤K≤1018,1≤d≤30000), denoting the required distance and the distance between every two adjacent checkpoints.

 

Output

For each test case, print a single line containing an integer, denoting the minimum distance.

 

Sample Input

12000 600 650 535 380

 

Sample Output

2165
Hint
The best path is 2-1-4-3-2.
 

 

题意：

给4个节点组成的环，求从点2出发，回到点2距离不小于k的最短路

思路：

这个回路肯定是由一个或者多个环组成的。一个环可以分成两个环x+ny的形式，n是y环的个数。而需要求解的环也可以分成这样两个环。

举例说，我们需要求解一个不小于1000的环，我们走2->1->2可以得到大小为30的环。如果我们能找到另一个环，大小为10，就能得到30*33+10=1000的环。但是这样的环不一定存在，如果存在大小为50的环，我们也可以得到30*32+50=1020的环。因为50>30，其实存在大小为20，50，80...的环都能得到的到1020的环。总而言之，我们只需要找是否存在环的大小%30等于0～29的环即可。

枚举量和y环的大小相关，所以我们要找最小的环。

具体看代码：

////  main.cpp//  1005////  Created by zc on 2017/8/25.//  Copyright © 2017年 zc. All rights reserved.//#include <iostream>#include<cstdio>#include<cstring>#include<cmath>#include<queue>#include<algorithm>#define ll long longusing namespace std;const int N= 66000;ll k,mp[5][5],ans,m,d[5][N];int v[5][N];void spfa(){    memset(d,0x3f,sizeof(d));    memset(v,0,sizeof(v));    queue<pair<int,ll> >q;    d[2][0]=0;    v[2][0]=1;    q.push(make_pair(2,0));    while(!q.empty())    {        auto t=q.front();        q.pop();        int u=t.first;        ll c=t.second;        v[u][c%m]=1;        if(u==2)        {            if(c>k)                ans=min(ans,c);            else                ans=min(ans,c+((k-c-1)/m+1)*m);        }        for(int i=-1;i<2;i+=2)        {            int p=(u+i+3)%4+1;            ll w=c+mp[u][p];            if(d[p][w%m]>w)            {                d[p][w%m]=w;                if(v[p][w%m]==0)                {                    v[p][w%m]=1;                    q.push(make_pair(p,w));                }            }        }    }}int main(int argc, const char * argv[]) {    int T;    scanf("%d",&T);    while(T--)    {        scanf("%lld",&k);        for(int i=1;i<=4;i++)        {            int j=(i%4)+1;            scanf("%lld",&mp[i][j]);            mp[j][i]=mp[i][j];        }        m=2*min(mp[2][1],mp[2][3]);        ans=((k-1)/m+1)*m;        spfa();        printf("%lld\n",ans);    }}