G. Bicycles

合集 - codeforces(57)

1.C. Insert and Equalize2023-12-062.C. Removal of Unattractive Pairs2023-12-063.D. Jumping Through Segments2023-12-084.E. Good Triples2023-12-085.F. Shift and Reverse2023-12-086.D. Yet Another Monster Fight2023-12-117.A. Constructive Problems2023-12-188.C. Game with Multiset2023-12-199.A. Rating Increase2023-12-1910.B. Swap and Delete2023-12-1911.A. Problemsolving Log2023-12-2012.B. Preparing for the Contest2023-12-2013.C. Quests2023-12-2014.D. Three Activities2023-12-2015.E2. Game with Marbles (Hard Version)2023-12-2016.cf刷题有感2023-12-2017.A. Anonymous Informant2023-12-2118.A. Forked!2023-12-2319.B. Make Almost Equal With Mod2023-12-2720.C. Heavy Intervals2023-12-2721.D. Split Plus K2023-12-2822.A. 20232023-12-3123.B. Two Divisors2023-12-3124.C. Training Before the Olympiad2023-12-3125.D. Mathematical Problem2024-01-0226.F. Greetings2024-01-0227.C. Partitioning the Array2024-01-15

28.G. Bicycles2024-01-16

29.E. Eat the Chip2024-01-1730.G. Lights2024-01-1731.D. Array Repetition2024-01-1732.D. Berserk Monsters2024-01-2233.E. Increasing Subsequences2024-01-2334.D. Very Different Array2024-01-2435.G. Mischievous Shooter2024-01-2536.B. Plus-Minus Split2024-01-2737.B. A Balanced Problemset2024-01-2838.C. Did We Get Everything Covered2024-01-2839.D. Find the Different Ones!2024-02-0840.C. Grouping Increases2024-02-1041.D. Good Trip2024-02-1242.C. Physical Education Lesson2024-02-1243.E. Final Countdown2024-02-1844.D. Divisible Pairs2024-02-1945.G. Vlad and Trouble at MIT2024-02-2046.A. Brick Wall2024-02-2847.B. Minimize Inversions2024-02-2848.C. XOR-distance2024-02-2849.A. Moving Chips2024-02-2850.B. Monsters Attack!2024-02-2851.C. Find B2024-02-2852.D. Slimes2024-02-2853.C. Turtle Fingers: Count the Values of k2024-02-2854.D. Turtle Tenacity: Continual Mods2024-02-2855.D. Vlad and Division2024-02-2856.C. LR-remainders2024-02-2857.D. Lonely Mountain Dungeons2024-03-12

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原题链接

题记，一道思考加编写加优化耗时2h的题

1.核心：抵达终点的路途中，如果换自行车，一定是换一辆速度系数更小的车
2.从速度系数最小的城市出发，到达终点的cost等于其系数乘上到达终点的最小距离
3.从速度系数第二小的城市出发，到达终点的最小值一定是直接往终点走和先去速度系数最小的城市之后再往终点走中的最小值
4.以此类推即可，细节我放在了code里，最细节的一点是这并不是一个多源最短路，而是单源最短路，在每个点作dijk的耗时比整体作spfa快很多

code

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

struct edge
{
    ll to;
    ll len;
};
vector<edge> G[1005];
void add(ll x,ll y,ll w)
{
    G[x].push_back({y,w});
    G[y].push_back({x,w});
}//以上操作为存边

struct unit
{
    ll speed;//这个城市自行车的速度系数
    ll id;
    ll cons;
}city[1005];
ll dis[1005];//代表某个点到剩下其余点的最小值
bool cmp(unit a,unit b){return a.speed<b.speed;}
struct fresh
{
    ll to;
    ll val;
    bool operator <(const fresh &b) const{return b.val<val;}//由于二元且升序，所以需要重载运算符
};

void dijk(ll who)//单源最短路
{
    memset(dis,0x3f3f3f,sizeof dis);
    priority_queue<fresh> q;
    q.push({who,0});
    while(q.size())
    {
        ll now=q.top().to;
        ll val=q.top().val;
        q.pop();
        if(dis[now]<=val)continue;
        dis[now]=val;
        for(int i=0;i<G[now].size();i++)
        {
            ll next=G[now][i].to;
            ll v=G[now][i].len+val;
            if(v<dis[next])  q.push({next,v});
        }
    }
}
int main()
{
    ios::sync_with_stdio(0);cin.tie(0);
    ll t;
    cin>>t;
    while(t--)
    {
        ll n,m;
        cin>>n>>m;
        for(int i=1;i<=n;i++)G[i].clear();
        for(ll i=1;i<=m;i++)
        {
            ll x,y,w;
            cin>>x>>y>>w;
            add(x,y,w);
        }

        for(ll i=1;i<=n;i++)
        {
            cin>>city[i].speed;
            city[i].id=i;
        }
        sort(city+1,city+n,cmp);//按速度排序，速度快的在前面，也就是速度系数小的在前面
        dijk(city[1].id);
        city[1].cons=city[1].speed*dis[n];
        ll ans=city[1].cons;//如果没有进入循环代表只有两个值，直接走就行了
        for(ll i=2;i<n;i++)
        {
            dijk(city[i].id);//由于我只需要知道现在这个点到其他点的最小值，并不需要知道所有点之间的距离，我所需要的也只是其他点出发到达n点的最小值
            city[i].cons=city[i].speed*dis[n];//代表直接走
            for(ll j=1;j<i;j++) city[i].cons=min(city[i].cons,city[i].speed*dis[city[j].id]+city[j].cons);//换自行车一定是换一辆速度比自己快的自行车
            if(city[i].id==1) ans=city[i].cons;
        }
        cout<<ans<<endl;
    }
    return 0;
}