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牛客小白月赛102-D-最短？路径

Posted on Edited on

题目大意

给定一个图，边有权值。在节点上每经过 k 条边必须休息一次（或者说每走 k 步必须停顿）。求从起点到终点的最短时间（或路径权值和），考虑休息带来的限制/代价。

AC代码

参考了官方题解思路，就是在分层图上跑最短路

image

相比于TLE代码，只加了个这个优化

while(!q.empty() && dist[n+1]==INF)

如果dist[n+1]已经确定，那就不用再跑了。

然后代码实现上有比较多的细节，主要是在初始化方面，比如说vis[i][0]也要清理什么的

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// https://ac.nowcoder.com/acm/contest/91355/D
#include <iostream>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long ll;
const ll N=2e5+10,INF=1e18+7;
ll T,n,m,k,a[N],u,v,dist[N];
vector<ll> e[N];
bool vis[N][11];
struct edge {
    ll node,dis,nrest; // nrest 为多久没休息了
};
struct cmp {
    bool operator()(edge a,edge b) {
        if(a.dis!=b.dis)
            return a.dis>b.dis;
        return a.node<b.node;
    }
};
priority_queue<edge,vector<edge>,cmp > q;
edge make_edge(ll node,ll dis,ll nrest) {
    edge temp;
    temp.node=node;temp.dis=dis;temp.nrest=nrest;
    return temp;
}
void dijkstra(int t) {
    q.push(make_edge(1,0,0));
    while(!q.empty() && dist[n+1]==INF) {
        ll node=q.top().node,dis=q.top().dis,nrest=q.top().nrest;
        q.pop();
        if(vis[node][nrest])    // 之前访问过，弹出
            continue;
        vis[node][nrest]=true;
        dist[node]=min(dis,dist[node]);
        for(int i=0;i<e[node].size();i++) {
            if(nrest>=k)
                q.push(make_edge(e[node][i],dis+a[node],0));
            else {
                q.push(make_edge(e[node][i],dis+a[node],0));
                q.push(make_edge(e[node][i],dis+1,nrest+1));
            }
        }
    }
}
void clear() {
    priority_queue<edge,vector<edge>,cmp > empty;
    swap(empty,q);
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    cin>>T;
    for(int _=1;_<=T;_++) {
        cin>>n>>m>>k;
        clear();
        for(int i=1;i<=n+1;i++)
            e[i].clear();
        for(int i=1;i<=n;i++) {
            cin>>a[i];
        }
        fill(&dist[1],&dist[n+2],INF);
        for(int i=1;i<=n+1;i++)
            for(int j=0;j<=k;j++)
                vis[i][j]=false;
        for(int i=1;i<=m;i++) {
            cin>>u>>v;
            e[u].push_back(v);
            e[v].push_back(u);
        }
        e[n].push_back(n+1);
        e[n+1].push_back(n);
        dijkstra(_);
        cout<<dist[n+1]<<endl;
    }
    return 0;
}
// TLE https://ac.nowcoder.com/acm/contest/view-submission?submissionId=72128029
// AC https://ac.nowcoder.com/acm/contest/view-submission?submissionId=72128343

心路历程（WA，TLE，MLE……）

过了40% TLE https://ac.nowcoder.com/acm/contest/view-submission?submissionId=72128029

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// https://ac.nowcoder.com/acm/contest/91355/D
#include <iostream>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
#include <queue>
using namespace std;
typedef long long ll;
const ll N=2e5+10,INF=1e18+7;
ll T,n,m,k,a[N],u,v,dist[N];
vector<ll> e[N];
bool vis[N][11];
struct edge {
    ll node,dis,nrest; // nrest 为多久没休息了
};
struct cmp {
    bool operator()(edge a,edge b) {
        if(a.dis!=b.dis)
            return a.dis>b.dis;
        return a.node<b.node;
    }
};
priority_queue<edge,vector<edge>,cmp > q;
edge make_edge(ll node,ll dis,ll nrest) {
    edge temp;
    temp.node=node;temp.dis=dis;temp.nrest=nrest;
    return temp;
}
void dijkstra(int t) {
    q.push(make_edge(1,0,0));
    while(!q.empty()) {
        ll node=q.top().node,dis=q.top().dis,nrest=q.top().nrest;
        q.pop();
        if(vis[node][nrest])    // 之前访问过，弹出
            continue;
        vis[node][nrest]=true;
        dist[node]=min(dis,dist[node]);
        for(int i=0;i<e[node].size();i++) {
            if(nrest>=k)
                q.push(make_edge(e[node][i],dis+a[node],0));
            else {
                q.push(make_edge(e[node][i],dis+a[node],0));
                q.push(make_edge(e[node][i],dis+1,nrest+1));
            }
        }
    }
}
void clear() {
    priority_queue<edge,vector<edge>,cmp > empty;
    swap(empty,q);
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    cin>>T;
    for(int _=1;_<=T;_++) {
        cin>>n>>m>>k;
        clear();
        for(int i=1;i<=n+1;i++)
            e[i].clear();
        for(int i=1;i<=n;i++) {
            cin>>a[i];
        }
        fill(&dist[1],&dist[n+2],INF);
        for(int i=1;i<=n+1;i++)
            for(int j=0;j<=k;j++)
                vis[i][j]=false;
        for(int i=1;i<=m;i++) {
            cin>>u>>v;
            e[u].push_back(v);
            e[v].push_back(u);
        }
        e[n].push_back(n+1);
        e[n+1].push_back(n);
        dijkstra(_);
        cout<<dist[n+1]<<endl;
    }
    return 0;
}
// TLE https://ac.nowcoder.com/acm/contest/view-submission?submissionId=72128029