bzoj 2763 [JLOI2011]飞行路线——分层图

Posted 2020-11-19 narh

题目：https://www.lydsy.com/JudgeOnline/problem.php?id=2763

分层图两种方法的练习。

1.把图分成k+1层，本层去上面一层的边免费。但空间时间都不算优秀。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#define ll long long
using namespace std;
const int N=1e4+5,M=5e4+5,K=15;
int n,m,k,head[N*K],xnt,s,t;
ll dis[N*K],ans=0x7fffffff;
bool vis[N*K];
struct Edge{
    int next,to,w;
    Edge(int n=0,int t=0,int w=0):next(n),to(t),w(w) {}
}edge[M*K<<2];
void add(int x,int y,int z)
{
    for(int fx=0;fx<=k*n;fx+=n)
    {
        edge[++xnt]=Edge(head[x+fx],y+fx,z);head[x+fx]=xnt;
        if(fx<k*n)
            edge[++xnt]=Edge(head[x+fx],y+fx+n,0),head[x+fx]=xnt;
    }
}
void dj()
{
    memset(dis,1,sizeof dis);dis[s]=0;
    priority_queue<pair<ll,int>,vector<pair<ll,int> >,greater<pair<ll,int> > > q;
    q.push(make_pair(0,s));
    while(q.size())
    {
        int u=q.top().second;q.pop();
        while(vis[u]&&q.size())u=q.top().second,q.pop();
        if(vis[u])break;vis[u]=1;
        for(int i=head[u],v;i;i=edge[i].next)
            if(dis[v=edge[i].to]>dis[u]+edge[i].w)
            {
                dis[v]=dis[u]+edge[i].w;q.push(make_pair(dis[v],v));
            }
    }
}
int main()
{
    scanf("%d%d%d%d%d",&n,&m,&k,&s,&t);
    int x,y,z;
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d%d",&x,&y,&z);
        add(x,y,z);add(y,x,z);
    }
    dj();
    for(int fx=0;fx<=k*n;fx+=n)ans=min(ans,dis[t+fx]);
    printf("%lld",ans);
    return 0;
}

View Code

2.设计dp状态，套在最短路上。就不用建边，从而省空间。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#define ll long long
using namespace std;
const int N=1e4+5,M=5e4+5,K=15;
int n,m,k,s,t,head[N],xnt;
ll ans=0x7fffffff,dp[N][K];
bool vis[N][K];
struct Edge{
    int next,to,w;
    Edge(int n=0,int t=0,int w=0):next(n),to(t),w(w) {}
}edge[M<<1];
void add(int x,int y,int z)
{
    edge[++xnt]=Edge(head[x],y,z);head[x]=xnt;
    edge[++xnt]=Edge(head[y],x,z);head[y]=xnt;
}
void spfa()
{
    memset(dp,1,sizeof dp);dp[s][0]=0;
    queue<pair<int,int> > q;q.push(make_pair(s,0));vis[s][0]=1;
    while(q.size())
    {
        int u=q.front().first,d=q.front().second;q.pop();
        vis[u][d]=0;
        for(int i=head[u],v;i;i=edge[i].next)
        {
            if(dp[v=edge[i].to][d]>dp[u][d]+edge[i].w)
            {
                dp[v][d]=dp[u][d]+edge[i].w;
                if(!vis[v][d])vis[v][d]=1,q.push(make_pair(v,d));
            }
            if(d<k&&dp[v][d+1]>dp[u][d])
            {
                dp[v][d+1]=dp[u][d];
                if(!vis[v][d+1])vis[v][d+1]=1,q.push(make_pair(v,d+1));
            }
        }
    }
}
int main()
{
    scanf("%d%d%d%d%d",&n,&m,&k,&s,&t);
    int x,y,z;
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d%d",&x,&y,&z);add(x,y,z);
    }
    spfa();
    for(int i=0;i<=k;i++)ans=min(ans,dp[t][i]);
    printf("%lld",ans);
    return 0;
}

View Code

　　上面的代码巨慢！用dj的话可以及时推出，所以会非常快！

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#define ll long long
using namespace std;
const int N=1e4+5,M=5e4+5,K=15;
int n,m,k,s,t,head[N],xnt;
ll ans=0x7fffffff,dp[N][K];
bool vis[N][K];
struct Edge{
    int next,to,w;
    Edge(int n=0,int t=0,int w=0):next(n),to(t),w(w) {}
}edge[M<<1];
struct Node{
    ll dis;int v,c;
    Node(ll d,int a,int b):dis(d),v(a),c(b) {}
    bool operator<(const Node k)const
    {
        return dis>k.dis;
    }
};
void add(int x,int y,int z)
{
    edge[++xnt]=Edge(head[x],y,z);head[x]=xnt;
    edge[++xnt]=Edge(head[y],x,z);head[y]=xnt;
}
void dj()
{
    memset(dp,1,sizeof dp);dp[s][0]=0;
    priority_queue<Node> q;
    q.push(Node(0,s,0));
    while(q.size())
    {
        ll x=q.top().dis,u=q.top().v,d=q.top().c;q.pop();
        while(vis[u][d]&&q.size())x=q.top().dis,u=q.top().v,d=q.top().c,q.pop();
        if(vis[u][d])break;vis[u][d]=1;
        if(u==t){ans=x;return;}
        for(int i=head[u],v;i;i=edge[i].next)
        {
            if(dp[v=edge[i].to][d]>x+edge[i].w)
            {
                dp[v][d]=x+edge[i].w;q.push(Node(dp[v][d],v,d));
            }
            if(d<k&&dp[v][d+1]>x)
            {
                dp[v][d+1]=x;q.push(Node(dp[v][d+1],v,d+1));
            }
        }
    }
}
int main()
{
    scanf("%d%d%d%d%d",&n,&m,&k,&s,&t);
    int x,y,z;
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d%d",&x,&y,&z);add(x,y,z);
    }
    dj();
    printf("%lld",ans);
    return 0;
}