P3350 [ZJOI2016] 旅行者

神仙题。
考虑分治，把当前矩阵较长的边平分成两半，把矩阵切开的那条线上的点更新在这个矩阵内的询问，然后继续分治。
正确性是显然的，时间复杂度为 \(T(n)=2T(\frac n2)+n^{\frac32}\log n\)，使用主定理得出 \(T(n) = \Theta(n^{\frac{3}{2}}\log n)\)，可以轻松通过。

代码

注意最短路写 \(SPFA\) 需要各种神秘的优化，所以我写的是 \(Dijkstra\)。

/*
Luogu P3350 [ZJOI2016] 旅行者
2026-03-30
*/
#include<bits/stdc++.h>
using namespace std;
namespace IO{
    template<typename T>
    inline void read(T&x){
        x=0;char c=getchar();bool f=0;
        while(!isdigit(c)) c=='-'?f=1:0,c=getchar();
        while(isdigit(c)) x=x*10+c-'0',c=getchar();
        f?x=-x:0;
    }
    template<typename T>
    inline void write(T x){
        if(x==0){putchar('0');return ;}
        x<0?x=-x,putchar('-'):0;short st[50],top=0;
        while(x) st[++top]=x%10,x/=10;
        while(top) putchar(st[top--]+'0');
    }
    inline void read(char&c){c=getchar();while(isspace(c)) c=getchar();}
    inline void write(char c){putchar(c);}
    inline void read(string&s){s.clear();char c;read(c);while(!isspace(c)&&~c) s+=c,c=getchar();}
    inline void write(string s){for(int i=0,len=s.size();i<len;i++) putchar(s[i]);}
    template<typename T>inline void write(T*x){while(*x) putchar(*(x++));}
    template<typename T,typename...T2> inline void read(T&x,T2&...y){read(x),read(y...);}
    template<typename T,typename...T2> inline void write(const T x,const T2...y){write(x),putchar(' '),write(y...),sizeof...(y)==1?putchar('\n'):0;}
}using namespace IO;
template<int maxn,int maxm>class LSQXX{
public:
    int head[maxn],nxt[maxm*2],to[maxm*2],val[maxm*2],cnt=0;
    void add(int u,int v,int w){nxt[++cnt]=head[u],to[cnt]=v,val[cnt]=w,head[u]=cnt;}
    void clear(){cnt=0;memset(head,0,sizeof(head));}
};
const int maxn=20010,inf=1000000000,maxq=100010;
int n,m,Q,ans[maxq],dis[maxn];
LSQXX<maxn,maxn*2>e;
priority_queue<pair<int,int>,vector<pair<int,int>>,greater<pair<int,int>>>qu;
struct Point{int x,y;};
struct Question{Point a,b;int ra,rb,id;}q[maxq],ls1[maxq],ls2[maxq];
void read(Point&a){read(a.x,a.y);}
int calc_id(int x,int y){return (x-1)*m+y;}
int calc_id(Point a){return (a.x-1)*m+a.y;}
void SPFA(int x,int y,int xx,int yy,int d){
    for(int i=x;i<=xx;i++) for(int j=y;j<=yy;j++) dis[calc_id(i,j)]=inf;
    dis[d]=0;qu.push({0,d});
    while(!qu.empty()){
        int u=qu.top().second;qu.pop();
        for(int i=e.head[u];i;i=e.nxt[i]){
            int v=e.to[i],z=e.val[i];
            if(dis[v]>dis[u]+z){
                dis[v]=dis[u]+z;
                qu.push({dis[v],v});
            }
        }
    }
}
void clear(int x,int y,int xx,int yy){for(int i=x;i<=xx;i++) for(int j=y;j<=yy;j++) dis[calc_id(i,j)]=0;}
void solve(int x,int y,int xx,int yy,int l,int r){
    if(x==xx&&y==yy){for(int i=l;i<=r;i++) ans[q[i].id]=0;return ;}
    if(xx-x>yy-y){
        int mid=x+xx>>1,cnt_ls1=0,cnt_ls2=0;
        for(int i=l;i<=r;i++){
            if(x<=q[i].a.x&&q[i].a.x<=mid&&x<=q[i].b.x&&q[i].b.x<=mid) ls1[++cnt_ls1]=q[i];
            if(mid+1<=q[i].a.x&&q[i].a.x<=xx&&mid+1<=q[i].b.x&&q[i].b.x<=xx) ls2[++cnt_ls2]=q[i];
        }
        for(int i=y;i<=yy;i++){
            SPFA(x,y,xx,yy,calc_id(mid,i));
            for(int j=l;j<=r;j++) ans[q[j].id]=min(ans[q[j].id],dis[q[j].ra]+dis[q[j].rb]);
            clear(x,y,xx,yy);
        }
        for(int i=l;i<=l+cnt_ls1-1;i++) q[i]=ls1[i-l+1];
        for(int i=l+cnt_ls1;i<=l+cnt_ls1+cnt_ls2-1;i++) q[i]=ls2[i-l-cnt_ls1+1];
        solve(x,y,mid,yy,l,l+cnt_ls1-1),solve(mid+1,y,xx,yy,l+cnt_ls1,l+cnt_ls1+cnt_ls2-1);
        return ;
    }
    int mid=y+yy>>1,cnt_ls1=0,cnt_ls2=0;
    for(int i=l;i<=r;i++){
        if(y<=q[i].a.y&&q[i].a.y<=mid&&y<=q[i].b.y&&q[i].b.y<=mid) ls1[++cnt_ls1]=q[i];
        if(mid+1<=q[i].a.y&&q[i].a.y<=yy&&mid+1<=q[i].b.y&&q[i].b.y<=yy) ls2[++cnt_ls2]=q[i];
    }
    for(int i=x;i<=xx;i++){
        SPFA(x,y,xx,yy,calc_id(i,mid));
        for(int j=l;j<=r;j++) ans[q[j].id]=min(ans[q[j].id],dis[q[j].ra]+dis[q[j].rb]);
        clear(x,y,xx,yy);
    }
    for(int i=l;i<=l+cnt_ls1-1;i++) q[i]=ls1[i-l+1];
    for(int i=l+cnt_ls1;i<=l+cnt_ls1+cnt_ls2-1;i++) q[i]=ls2[i-l-cnt_ls1+1];
    solve(x,y,xx,mid,l,l+cnt_ls1-1),solve(x,mid+1,xx,yy,l+cnt_ls1,l+cnt_ls1+cnt_ls2-1);
}
signed main(){
    read(n,m);
    for(int i=1;i<=n;i++) for(int j=1;j<m;j++){
        int u,v,w;read(w);
        u=calc_id(i,j),v=calc_id(i,j+1);
        e.add(u,v,w),e.add(v,u,w);
    }
    for(int i=1;i<n;i++) for(int j=1;j<=m;j++){
        int u,v,w;read(w);
        u=calc_id(i,j),v=calc_id(i+1,j);
        e.add(u,v,w),e.add(v,u,w);
    }
    read(Q);
    for(int i=1;i<=Q;i++) read(q[i].a,q[i].b),q[i].id=i,q[i].ra=calc_id(q[i].a),q[i].rb=calc_id(q[i].b);
    memset(ans,0x3f,sizeof(ans));
    solve(1,1,n,m,1,Q);
    for(int i=1;i<=Q;i++) write(ans[i]),write("\n");
    return 0;
}