BZOJ 4456 [Zjoi2016]旅行者

分治，solve(x1,x2,y1,y2,ask[])表示在某个矩形的范围内解决ask内的询问

具体方法是找x,y中较短的一维，假设是x,则设mid=(x1+x2)/2,然后在mid那根线上每个点来一次Dij，每个点i做完之后对于ask中每个询问更新一次（可提供更新的答案的意义是经过i的最短路是多少），然后对于两个点分别在mid两边的就不再递归，否则继续分治下去。

好像以前做过一道类似的noip集训分治题目是判连通性orz，然后这题很明显的看出来了分治，只不过最开始以为只沿一维分割即可后来发现需要用短的一维分割，大概是为了减少跑Dij的次数。

代码vector常数比较大

//QWsin
#include<cmath>
#include<queue>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int INF=1<<28;
const int maxn=20000+10;
const int maxq=100000+10;
const int maxm=100000+10;
inline int read(){
int ret=0;char ch=getchar();
while(ch<'0'||ch>'9') ch=getchar();
for(;ch>='0'&&ch<='9';ch=getchar()) ret=ret*10+ch-'0';
return ret;
}

int n,m,ans[maxq];
inline int ID(int x,int y){return (x-1)*m+y;}

int first[maxn],next[maxm],ecnt;
struct Edge{int u,v,w;Edge(int u=0,int v=0,int w=0):u(u),v(v),w(w){}}e[maxm];
inline void add_edge(int u,int v,int w){
next[ecnt]=first[u];first[u]=ecnt;e[ecnt++]=Edge(u,v,w);
next[ecnt]=first[v];first[v]=ecnt;e[ecnt++]=Edge(v,u,w);
}

struct Ask{
int x1,y1,x2,y2,id;
Ask(int x1,int y1,int x2,int y2,int id)
:x1(x1),y1(y1),x2(x2),y2(y2),id(id){}
};

struct Node{
int u,dis;
Node(int u,int dis):u(u),dis(dis){}
inline bool operator < (const Node &rhs)const{
return dis>rhs.dis;
}
};

inline int getx(int v){return (v-1)/m+1;}
inline int gety(int v){return (v-1)%m+1;}

int dis[maxn],vis[maxn];

priority_queue<Node>q;
inline void Dij(int s,const int &x1,const int &x2,const int &y1,const int &y2)
{
for(int i=x1;i<=x2;++i)
for(int j=y1;j<=y2;++j)
dis[ID(i,j)]=INF,vis[ID(i,j)]=0;

dis[s]=0;
q.push(Node(s,0));
while(!q.empty())
{
Node t=q.top();q.pop();
int u=t.u;
if(vis[u]) continue;
vis[u]=1;

for(int i=first[u];i!=-1;i=next[i])
{
int v=e[i].v,x=getx(v),y=gety(v);
if(x<x1||x>x2||y<y1||y>y2) continue;
if(dis[v] > dis[u] + e[i].w){
dis[v]=dis[u]+e[i].w;
q.push(Node(v,dis[v]));
}
}
}
}

void solve(int x1,int x2,int y1,int y2,vector<Ask>&ask)
{
if(!ask.size()) return ;

vector<Ask>askl,askr;
int sz=ask.size();

if(x2-x1 > y2-y1)
{
int mid=(x1+x2)>>1;
for(int i=y1;i<=y2;++i)
{
Dij(ID(mid,i),x1,x2,y1,y2);
for(int i=0;i<sz;++i)
{
Ask t=ask[i];
ans[t.id]=min(ans[t.id],dis[ID(t.x1,t.y1)]+dis[ID(t.x2,t.y2)]);
}
}

for(int i=0;i<sz;++i)
if(ask[i].x1<mid&&ask[i].x2<mid) askl.push_back(ask[i]);
else if(ask[i].x1>mid&&ask[i].x2>mid) askr.push_back(ask[i]);

solve(x1,mid-1,y1,y2,askl);
solve(mid+1,x2,y1,y2,askr);
}
else
{
int mid=(y1+y2)>>1;
for(int i=x1;i<=x2;++i){
Dij(ID(i,mid),x1,x2,y1,y2);
for(int i=0;i<sz;++i)
{
Ask t=ask[i];
ans[t.id]=min(ans[t.id],dis[ID(t.x1,t.y1)]+dis[ID(t.x2,t.y2)]);
}
}

for(int i=0;i<sz;++i)
if(ask[i].y1<mid&&ask[i].y2<mid) askl.push_back(ask[i]);
else if(ask[i].y1>mid&&ask[i].y2>mid) askr.push_back(ask[i]);

solve(x1,x2,y1,mid-1,askl);
solve(x1,x2,mid+1,y2,askr);
}
}

int main()
{
cin>>n>>m;
memset(first,-1,sizeof first);

for(int i=1;i<=n;++i)
for(int j=1;j<m;++j)
add_edge(ID(i,j),ID(i,j+1),read());

for(int i=1;i<n;++i)
for(int j=1;j<=m;++j)
add_edge(ID(i,j),ID(i+1,j),read());

vector<Ask>ask;

int Q,x1,y1,x2,y2;cin>>Q;
for(int i=1;i<=Q;++i){
ans[i]=INF;

x1=read();y1=read();x2=read();y2=read();
ask.push_back(Ask(x1,y1,x2,y2,i));
}

solve(1,n,1,m,ask);

for(int i=1;i<=Q;++i) printf("%d\n",ans[i]);
return 0;
}