luogu 5468 [NOI2019]回家路线 最短路/暴力

这道题的斜率优化也比较简单呀

想写一个 70 pts 算法，结果数据水，直接就切了

最短路： 

// luogu-judger-enable-o2
#include<bits/stdc++.h>
using namespace std;
#define setIO(s) freopen(s".in","r",stdin) 
#define maxn 300000 
#define ll long long 
struct Node { int x,y,p,q;  }nd[maxn];  
vector<int>G1[maxn],G2[maxn];    
map<int,int>date1[maxn],date2[maxn];         
int n,m,A,B,C,cnt1,edges,cnt2; 
int hd[maxn*10],to[maxn*10],nex[maxn*10]; 
ll val[maxn*10]; 
inline void addedge(int u,int v,ll c)  
{   
    nex[++edges]=hd[u], hd[u]=edges, to[edges]=v,val[edges]=c;  
} 
inline ll calc(ll u) { return 1ll*A*u*u + 1ll*B*u + 1ll*C; }     
int S,T;  
int inq[maxn*10]; 
ll dis[maxn*10]; 
queue<int>Q; 
inline void spfa() 
{ 
    for(int i=1;i<maxn*10;++i) dis[i]=1000000000000; 
    Q.push(S); 
    inq[S]=1;
    dis[S]=0; 
    while(!Q.empty())
    {
        int u=Q.front(); Q.pop();  
        inq[u]=0; 
        for(int i=hd[u];i;i=nex[i]) 
        {
            int v=to[i];  
            if(dis[u]+val[i] < dis[v]) 
            {
                dis[v]=dis[u] + val[i]; 
                if(!inq[v]) 
                {
                    inq[v]=1; 
                    Q.push(v); 
                }
            }
        }
    } 
    printf("%lld
",dis[T]); 
}
int main()
{ 
    int i,j; 
    // setIO("input");        
    scanf("%d%d%d%d%d",&n,&m,&A,&B,&C);  
    for(i=1;i<=m;++i) 
    {
        scanf("%d%d%d%d",&nd[i].x,&nd[i].y,&nd[i].p,&nd[i].q);   
        G1[nd[i].y].push_back(nd[i].q);  
        G2[nd[i].x].push_back(nd[i].p); 
        if(!date2[nd[i].x][nd[i].p]) date2[nd[i].x][nd[i].p]=++cnt2;  
        if(!date1[nd[i].y][nd[i].q]) date1[nd[i].y][nd[i].q]=++cnt1;           
    }    
    for(i=1;i<=m;++i) 
    {
        addedge(date2[nd[i].x][nd[i].p], date1[nd[i].y][nd[i].q] + cnt2, 0); 
    }
    int k=0;     
    for(i=1;i<=n;++i) 
    {   
        if(!G1[i].size() || !G2[i].size()) continue;   
        for(j=0;j<G1[i].size();++j) 
        {
            for(k=0;k<G2[i].size();++k) 
            { 
                if(G1[i][j]>G2[i][k]) continue;    
                addedge(date1[i][G1[i][j]] + cnt2, date2[i][G2[i][k]], calc(G2[i][k] - G1[i][j])); 
            }
        }      
    }   
    S=0,T=cnt1 + cnt2 +100;   
    for(i=0;i<=1000;++i)  if(date2[1][i]) addedge(S, date2[1][i], calc(i));   
    for(i=0;i<=1000;++i)  if(date1[n][i]) addedge(date1[n][i] + cnt2, T, i);   
    spfa();        
    return 0; 
}

　　

$O(mt)$ DP (网上说这个做法在现场能切题)

#include<bits/stdc++.h>
#define setIO(s) freopen(s".in","r",stdin) 
#define maxn 300001 
#define inf 1000000000
using namespace std;
int n,m,A,B,C; 
int f[maxn][1002];
int calc(int t) 
{
    return A*t*t+B*t+C; 
}  
struct Node{ int x,y,p,q; }nd[maxn];
bool cmp(Node a, Node b)
{
    return a.p < b.p; 
}
int main()
{
    //    setIO("input");  
    int i,j; 
    scanf("%d%d%d%d%d",&n,&m,&A,&B,&C); 
    for(i=1;i<=m;++i) 
    {
        scanf("%d%d%d%d",&nd[i].x,&nd[i].y,&nd[i].p,&nd[i].q); 
    }
    sort(nd+1,nd+1+m,cmp);    
    for(i=0;i<=n;++i) for(j=0;j<=1000;++j) f[i][j]=inf;         
    f[1][0]=0;  
    for(i=1;i<=m;++i) 
    {
        for(j=0;j<=nd[i].p;++j) 
        {
            if(f[nd[i].x][j]==inf) continue;     
            f[nd[i].y][nd[i].q] = min(f[nd[i].y][nd[i].q], f[nd[i].x][j] + calc(nd[i].p - j));  
        }
    }
    int ans=inf; 
    for(i=0;i<=1000;++i) 
    {
        ans=min(ans, f[n][i] + i); 
    }
    printf("%d
",ans); 
    return 0; 
}