loj2090. 「ZJOI2016」旅行者
链接
loj
思路
\((l,mid)(mid+1,r)\).考虑跨过mid的贡献。
假设选的中间那条线的点为gzy,贡献为\(dis(x,gzy)+dis(gzy,y)\)
那就计算n遍最短路,一次分治为\(n^2mlog{nm}\)
设S=n*m.矩阵的长度是不定的，每次取最长的边进行分治是最好的，n最坏为\(\sqrt{n}\)。
\(f(n)=2*f(\frac{n}{2})+S\sqrt{S}logS。所以总的复杂度就是\)\(S\sqrt{S}logS\)
都在同侧的也需要跨一跨
代码
#include <bits/stdc++.h>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
using namespace std;
const int _=1e5+7,INF=0x3f3f3f3f;
int read() {int x=0,f=1;char s=getchar();for(;s>'9'||s<'0';s=getchar()) if(s=='-') f=-1;for(;s>='0'&&s<='9';s=getchar()) x=x*10+s-'0';return x*f;
}
int n,m,q,ans[_],vis[_];
struct node {int x,y,X,Y,u,v,id;
}Q[_],tmp[_];
bool cmp(node a,node b) {return a.id<b.id;}
struct edge {int v,nxt,q;}e[_];
int head[_],tot;
void add(int u,int v,int q) {e[++tot].v=v;e[tot].q=q;e[tot].nxt=head[u];head[u]=tot;
}
int id(int x,int y) {return (x-1)*m+y;}
struct T_T {int u,val;T_T(int a=0,int b=0) {u=a,val=b;}bool operator < (const T_T &b) const {return val>b.val;}
};
int dis[_];
void dij(int S) {dis[S]=0;priority_queue<T_T> q;q.push(T_T(S,0));while(!q.empty()) {T_T u=q.top();q.pop();if(dis[u.u]!=u.val) continue;for(int i=head[u.u];i;i=e[i].nxt) {int v=e[i].v;if(vis[v]&&dis[v]>u.val+e[i].q) {dis[v]=u.val+e[i].q;q.push(T_T(v,dis[v]));}}}
}
void solve(int x,int y,int X,int Y,int l,int r) {if(l>r) return;if(x==X&&y==Y) {for(int i=l;i<=r;++i) ans[Q[i].id]=0;return;}if(Y-y>X-x) {int mid=(Y+y)>>1;FOR(i,x,X) {FOR(j,x,X) FOR(k,y,Y) dis[id(j,k)]=INF;dij(id(i,mid));FOR(j,l,r)ans[Q[j].id]=min(dis[Q[j].u]+dis[Q[j].v],ans[Q[j].id]);}FOR(i,x,X) vis[id(i,mid)]=0;int p=l,q=r;FOR(i,l,r) {if(max(Q[i].y,Q[i].Y)<=mid) tmp[p++]=Q[i];if(min(Q[i].y,Q[i].Y)>mid) tmp[q--]=Q[i];}FOR(i,l,r) Q[i]=tmp[i];solve(x,y,X,mid,l,p-1);solve(x,mid+1,X,Y,q+1,r);} else {int mid=(X+x)>>1;FOR(i,y,Y) {FOR(j,x,X) FOR(k,y,Y) dis[id(j,k)]=INF;dij(id(mid,i));FOR(j,l,r)ans[Q[j].id]=min(dis[Q[j].u]+dis[Q[j].v],ans[Q[j].id]);}FOR(i,y,Y) vis[id(mid,i)]=0;int p=l,q=r;FOR(i,l,r) {if(max(Q[i].x,Q[i].X)<=mid) tmp[p++]=Q[i];if(min(Q[i].x,Q[i].X)>mid) tmp[q--]=Q[i];}FOR(i,l,r) Q[i]=tmp[i];solve(x,y,mid,Y,l,p-1);solve(mid+1,y,X,Y,q+1,r);}
}
int main() {n=read(),m=read();FOR(i,1,n) FOR(j,1,m-1) {int val=read();add(id(i,j),id(i,j+1),val);add(id(i,j+1),id(i,j),val);}FOR(i,1,n-1) FOR(j,1,m) {int val=read();add(id(i,j),id(i+1,j),val);add(id(i+1,j),id(i,j),val);}FOR(i,1,n) FOR(j,1,m) vis[id(i,j)]=1;q=read();FOR(i,1,q) {Q[i].x=read(),Q[i].y=read(),Q[i].u=id(Q[i].x,Q[i].y);Q[i].X=read(),Q[i].Y=read(),Q[i].v=id(Q[i].X,Q[i].Y);Q[i].id=i,ans[i]=INF;}solve(1,1,n,m,1,q);FOR(i,1,q) printf("%d\n",ans[i]);return 0;
}
