BZOJ4456 ZJOI2016旅行者（分治+最短路）

　　感觉比较套路，每次在长边中轴线处切一刀，求出切割线上的点对矩形内所有点的单源最短路径，以此更新每个询问，递归处理更小的矩形。因为若起点终点跨过中轴线是肯定要经过的，而不跨过中轴线的则可以选择是否经过中轴线，若不经过一定就在矩形的某一半了。复杂度O((nm)^1.5log(nm))，不太会证。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
#define N 20010
#define M 200
#define Q 100010
int n,m,u,op,a[N*M][4],dis[M][N],ans[Q];
struct data{int sx,sy,tx,ty,i;
}q[Q],tmp[Q];
int wx[4]={-1,0,1,0},wy[4]={0,1,0,-1};
int trans(int x,int y){return (x-1)*m+y;}
namespace shortestpath
{
    int p[N],t;bool flag[N];
    struct data{int to,nxt,len;}edge[N<<2];
    struct data2
    {
        int x,d;
        bool operator <(const data2&a) const
        {
            return d>a.d;
        }
    };
    priority_queue<data2> q;
    void addedge(int x,int y,int z){t++;edge[t].to=y,edge[t].nxt=p[x],edge[t].len=z,p[x]=t;}
    void dijkstra(int x,int y,int k)
    {
        while (!q.empty()) q.pop();
        memset(dis[k],42,sizeof(dis[k]));dis[k][trans(x,y)]=0;q.push((data2){trans(x,y),0});
        memset(flag,0,sizeof(flag));
        while (1)
        {
            while (!q.empty()&&flag[q.top().x]) q.pop();
            if (q.empty()) break;
            data2 v=q.top();q.pop();
            flag[v.x]=1;
            for (int j=p[v.x];j;j=edge[j].nxt)
            if (v.d+edge[j].len<dis[k][edge[j].to])
            {
                dis[k][edge[j].to]=v.d+edge[j].len;
                q.push((data2){edge[j].to,dis[k][edge[j].to]});
            }
        }
    }
    void make(int u,int d,int l,int r)
    {
        t=0;
        for (int i=u;i<=d;i++)
            for (int j=l;j<=r;j++)
            p[trans(i,j)]=0;
        for (int i=u;i<=d;i++)
            for (int j=l;j<=r;j++)
                for (int k=0;k<4;k++)
                if (i+wx[k]>=u&&i+wx[k]<=d&&j+wy[k]>=l&&j+wy[k]<=r)
                addedge(trans(i,j),trans(i+wx[k],j+wy[k]),a[trans(i,j)][k]);
    }
}
void solve(int u,int d,int l,int r,int x,int y)
{
    if (u>d||l>r||x>y) return;
    shortestpath::make(u,d,l,r);
    if (d-u<=r-l)
    {
        int mid=l+r>>1,s=x-1,t=y+1;
        for (int i=u;i<=d;i++) shortestpath::dijkstra(i,mid,i-u+1);
        for (int i=x;i<=y;i++)
        {
            for (int j=1;j<=d-u+1;j++)
            ans[q[i].i]=min(ans[q[i].i],dis[j][trans(q[i].sx,q[i].sy)]+dis[j][trans(q[i].tx,q[i].ty)]);
            if (max(q[i].sy,q[i].ty)<mid) tmp[++s]=q[i];
            else if (min(q[i].sy,q[i].ty)>mid) tmp[--t]=q[i];
        }
        for (int i=x;i<=s;i++) q[i]=tmp[i];
        for (int i=t;i<=y;i++) q[i]=tmp[i];
        solve(u,d,l,mid-1,x,s),
        solve(u,d,mid+1,r,t,y);
    } 
    else
    {
        int mid=u+d>>1,s=x-1,t=y+1;
        for (int i=l;i<=r;i++) shortestpath::dijkstra(mid,i,i-l+1);
        for (int i=x;i<=y;i++)
        {
            for (int j=1;j<=r-l+1;j++)
            ans[q[i].i]=min(ans[q[i].i],dis[j][trans(q[i].sx,q[i].sy)]+dis[j][trans(q[i].tx,q[i].ty)]);
            if (max(q[i].sx,q[i].tx)<mid) tmp[++s]=q[i];
            else if (min(q[i].sx,q[i].tx)>mid) tmp[--t]=q[i];
        }
        for (int i=x;i<=s;i++) q[i]=tmp[i];
        for (int i=t;i<=y;i++) q[i]=tmp[i];
        solve(u,mid-1,l,r,x,s),
        solve(mid+1,d,l,r,t,y);
    }
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj4456.in","r",stdin);
    freopen("bzoj4456.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read(),m=read();
    for (int i=1;i<=n;i++)
        for (int j=1;j<m;j++)
        a[trans(i,j)][1]=a[trans(i,j+1)][3]=read();
    for (int i=1;i<n;i++)
        for (int j=1;j<=m;j++)
        a[trans(i,j)][2]=a[trans(i+1,j)][0]=read();
    u=read();
    for (int i=1;i<=u;i++)
    q[i].sx=read(),q[i].sy=read(),q[i].tx=read(),q[i].ty=read(),q[i].i=i;
    memset(ans,42,sizeof(ans));
    solve(1,n,1,m,1,u);
    for (int i=1;i<=u;i++) printf("%d\n",ans[i]);
    return 0;
}