P9751 [CSP-J 2023] 旅游巴士

合集 - 图论(79)

1.F. Chat Screenshots2024-02-192.P1656 炸铁路2024-02-223.P1137 旅行计划2024-02-244.P2835 刻录光盘2024-02-245.P1197 [JSOI2008] 星球大战2024-02-246.P3388 【模板】割点（割顶）2024-02-247.P8435 【模板】点双连通分量2024-02-258.P8436 【模板】边双连通分量2024-02-259.P2860 [USACO06JAN] Redundant Paths G2024-02-2510.P1653 [USACO04DEC] Cow Ski Area G2024-02-2611.P3047 [USACO12FEB] Nearby Cows G2024-03-0412.P1894 [USACO4.2] 完美的牛栏The Perfect Stall2024-03-0613.P1550 [USACO08OCT] Watering Hole G2024-03-0614.P2330 [SCOI2005] 繁忙的都市2024-03-0615.P1525 [NOIP2010 提高组] 关押罪犯2024-03-0716.P1379 八数码难题2024-03-0717.P2746 [USACO5.3] 校园网Network of Schools2024-03-0818.P6121 [USACO16OPEN] Closing the Farm G2024-03-0819.P2341 [USACO03FALL / HAOI2006] 受欢迎的牛 G2024-03-0820.P2055 [ZJOI2009] 假期的宿舍2024-03-0821.P5905 【模板】全源最短路（Johnson）2024-03-1122.F. Microcycle2024-03-1223.G. Path Prefixes2024-03-1424.G. Rudolf and Subway2024-03-1525.C. Ehab and Path-etic MEXs2024-03-1726.A. String Transformation 12024-03-1727.D. Secret Passwords2024-03-1928.F. Maximum White Subtree2024-03-1929.P3478 [POI2008] STA-Station2024-03-1930.P1347 排序2024-03-2131.P1960 郁闷的记者2024-03-2132.E1. Weights Division (easy version)2024-03-2233.P5007 DDOSvoid 的疑惑2024-03-2534.P2850 [USACO06DEC] Wormholes G2024-03-2635.P1265 公路修建2024-03-2836.P1354 房间最短路问题2024-03-2937.P2168 [NOI2015] 荷马史诗2024-03-3038.P8306 【模板】字典树2024-03-3039.P1481 魔族密码2024-03-3040.P3128 [USACO15DEC] Max Flow P2024-04-0241.P5536 【XR-3】核心城市2024-04-0242.P5836 [USACO19DEC] Milk Visits S2024-04-0243.P3384 【模板】重链剖分/树链剖分2024-04-0344.P5960 【模板】差分约束2024-04-0445.P7771 【模板】欧拉路径2024-04-0546.六度分离2024-04-0847.整数区间2024-04-0848.F. Alex's whims2024-04-1249.J. 上学2024-05-1450.Game on Tree2024-05-1451.E. We Need More Bosses2024-05-1552.B. Omkar and Heavenly Tree2024-05-1653.B. Mahmoud and Ehab and the bipartiteness2024-05-1654.P1668 [USACO04DEC] Cleaning Shifts S2024-05-1755.P6154 游走2024-05-2056.P8655 [蓝桥杯 2017 国 B] 发现环2024-05-2457.P10298 [CCC 2024 S4] Painting Roads2024-05-2458.P9650 [SNCPC2019] Escape Plan2024-05-2759.P9327 [CCC 2023 S4] Minimum Cost Roads2024-05-2960.P9026 [CCC2021 S4] Daily Commute2024-05-2961.P8724 [蓝桥杯 2020 省 AB3] 限高杆2024-06-0362.P4878 [USACO05DEC] Layout G2024-06-0463.P5663 [CSP-J2019] 加工零件2024-06-0464.P2731 [USACO3.3] 骑马修栅栏 Riding the Fences2024-06-0865.I. Disks2024-06-1766.P1351 [NOIP2014 提高组] 联合权值2024-06-2667.B. Time Travel2024-06-2968.F. Minimum Maximum Distance2024-07-1469.A. Book2024-07-1970.P1407 [国家集训队] 稳定婚姻2024-07-2071.P1991 无线通讯网2024-07-2372.P4047 [JSOI2010] 部落划分2024-07-2373.P3275 [SCOI2011] 糖果2024-07-2374.P1989 无向图三元环计数2024-07-2775.P1967 [NOIP2013 提高组] 货车运输2024-07-3176.D. Vitaly and Cycle2024-08-0177.P10838 『FLA - I』庭中有奇树2024-08-07

78.P9751 [CSP-J 2023] 旅游巴士2024-08-11

79.D. Colored Portals2024-08-21

收起

原题链接

分析

很逆天的一道题

设 $dp[i][j]$ 为到达第 $i$ 个点的时刻 $t$ 且满足 $t\mathrm{mod}k=j$ 的最小 $t$

则有答案为 $dp[n][0]$

更新也很简单，设当前点为 $u$，当前时间为 $t$ 需要遍历的下一个点 $v$，则有 $dis[v][(t+1)\mathrm{\%}k]=dis[u][t\mathrm{\%}k]+1$

如果道路还没开放，我们可以原地等 $k$ 的倍数时间（等价于在起点等），因为我们只更新 $[(t+1)\mathrm{\%}k]$，其他的时间自然会有其他时刻来更新

这个设计非常巧妙，避开了 $a_i$ 的限制

code

#include<bits/stdc++.h>
using namespace std;
/*
#define int long long
#define double long double
#define lowbit(x) ((x)&(-x))
const int inf=1e18;
const int mod=1e9+7;

const int N=4e5;
int qpow(int a,int n)
{
    int res=1;
    while(n)
    {
        if(n&1) res=res*a%mod;
        a=a*a%mod;
        n>>=1;
    }
    return res;
}
int inv(int x)
{
    return qpow(x,mod-2);
}
int fa[2000005];
int finds(int now){return now==fa[now]?now:finds(fa[now]);}

vector<int> G[200005];

int dfn[200005],low[200005];
int cnt=0,num=0;
int in_st[200005]={0};
stack<int> st;
int belong[200005]={0};

void scc(int now,int fa)
{
    dfn[now]=++cnt;
    low[now]=dfn[now];
    in_st[now]=1;
    st.push(now);

    for(auto next:G[now])
    {
        if(next==fa) continue;

        if(!dfn[next])
        {
            scc(next,now);
            low[now]=min(low[now],low[next]);
        }
        else if(in_st[next])
        {
            low[now]=min(low[now],dfn[next]);
        }
    }

    if(low[now]==dfn[now])
    {
        int x;
        num++;
        do
        {
            x=st.top();
            st.pop();
            in_st[x]=0;
            belong[x]=num;
        }while(x!=now);
    }
}
vector<int> prime;
bool mark[200005]={0};
void shai()
{
    for(int i=2;i<=200000;i++)
    {
        if(!mark[i]) prime.push_back(i);

        for(auto it:prime)
        {
            if(it*i>200000) break;

            mark[it*i]=1;
            if(it%i==0) break;
        }
    }
}
*/

struct node
{
    int to,cost;
};
vector<node> G[10004];

struct fresh
{
    int place,mod,t;
    bool operator<(const fresh&c)const
    {
        return c.t<t;
    }
};


int dis[10005][105];
void solve()
{
    memset(dis,0x3f,sizeof dis);
    int n,m,k;
    cin>>n>>m>>k;

    for(int i=1;i<=m;i++)
    {
        int x,y,t;
        cin>>x>>y>>t;

        G[x].push_back({y,t});
    }


    priority_queue<fresh> q;
    q.push({1,0,0});

    while(q.size())
    {
        auto [now,mod,t]=q.top();
        q.pop();
        if(dis[now][mod]<=t) continue;

        dis[now][mod]=t;
        for(auto next:G[now])
        {
            auto [to,wait]=next;

            if(wait<=t)
            {
                q.push({to,(mod+1)%k,t+1});
            }
            else
            {
                int add=(wait-t)/k+((wait-t)%k!=0);
                q.push({to,(mod+1)%k,t+add*k+1});
            }
        }
    }

    if(dis[n][0]>1e9) cout<<-1;
    else cout<<dis[n][0];
}
signed main()
{
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int TT=1;
    //cin>>TT;
    while(TT--) solve();
    return 0;
}