BZOJ 4456 [Zjoi2016]旅行者

题解：分治

div(a,b,c,d,l,r)

表示处理在(a,b)(c,d)这个矩形内走，队列(l,r)中询问的答案

枚举较短中线上的点，求最短路

如果x,y不再同一侧，那么最短路一定经过中线，处理完毕

如果在同一侧，那么最短路可能经过也可能不经过中线，所以递归两边处理

犯过的SB错误：

把询问分组的时候把原数组覆盖了

横纵坐标不分

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int maxn=300009;
const int oo=1000000000;

int n,m,TT;
int downdist[maxn],rightdist[maxn];

int limx1,limy1,limx2,limy2;

inline int P(int x,int y){
    return (x-1)*m+y;
}
inline bool Isinrange(int u){
    int y=(u-1)%m+1;
    int x=(u-y)/m+1;
    return (x>=limx1)&&(x<=limx2)&&(y>=limy1)&&(y<=limy2);
}


int d[maxn];
int vis[maxn];
struct HeapNode{
    int v,mindist;
    HeapNode(int x){
        v=x;mindist=d[x];
    }
    bool operator < (const HeapNode &rhs) const{
        return mindist>rhs.mindist;
    }
};
priority_queue<HeapNode>q;

void Dijkstra(int s){
    for(int i=limx1;i<=limx2;++i){
        for(int j=limy1;j<=limy2;++j){
            d[P(i,j)]=oo;vis[P(i,j)]=0;
        }
    }
    
    d[s]=0;q.push(HeapNode(s));
    while(!q.empty()){
        HeapNode x=q.top();q.pop();
        int u=x.v;
        if(vis[u])continue;
        vis[u]=1;
        int v;
        v=u+m;
        if(v<=n*m){
            if(d[u]+downdist[u]<d[v]){
                d[v]=d[u]+downdist[u];
                q.push(HeapNode(v));
            }
        }
        v=u-m;
        if(v>0){
            if(d[u]+downdist[v]<d[v]){
                d[v]=d[u]+downdist[v];
                q.push(HeapNode(v));
            }
        }
        v=u+1;
        if((u%m!=0)){
            if(d[u]+rightdist[u]<d[v]){
                d[v]=d[u]+rightdist[u];
                q.push(HeapNode(v));
            }
        }
        v=u-1;
        if(((u-1)%m!=0)){
            if(d[u]+rightdist[v]<d[v]){
                d[v]=d[u]+rightdist[v];
                q.push(HeapNode(v));
            }
        }
    }
}

int qs[maxn];
int qx1[maxn],qy1[maxn],qx2[maxn],qy2[maxn];
int ans[maxn];
int tmpq[maxn];

void DivCon(int upx,int upy,int downx,int downy,int ll,int rr){
    if(ll>rr)return;
    int deltx=downx-upx;
    int delty=downy-upy;
    limx1=upx;limy1=upy;
    limx2=downx;limy2=downy;

    int p1=ll,p2=rr;
    if(deltx<=delty){
        int mid=(upy+downy)>>1;
        for(int i=upx;i<=downx;++i){
            Dijkstra(P(i,mid));
            for(int k=ll;k<=rr;++k){
                int t=qs[k];
                int a=P(qx1[t],qy1[t]);
                int b=P(qx2[t],qy2[t]);
                ans[t]=min(ans[t],d[a]+d[b]);
            }
        }
        for(int k=ll;k<=rr;++k){
            int t=qs[k];
            if(((qy1[t]<=mid)&&(qy2[t]>=mid))||((qy1[t]>=mid)&&(qy2[t]<=mid)))continue;
            if(qy1[t]<=mid){
                tmpq[p1]=t;++p1;
            }
            if(qy1[t]>=mid){
                tmpq[p2]=t;--p2;
            }
        }
        for(int i=ll;i<=rr;++i)qs[i]=tmpq[i];
        DivCon(upx,upy,downx,mid-1,ll,p1-1);
        DivCon(upx,mid+1,downx,downy,p2+1,rr);
    }else{
        int mid=(upx+downx)>>1;
        for(int i=upy;i<=downy;++i){
            Dijkstra(P(mid,i));
            for(int k=ll;k<=rr;++k){
                int t=qs[k];
                int a=P(qx1[t],qy1[t]);
                int b=P(qx2[t],qy2[t]);
                ans[t]=min(ans[t],d[a]+d[b]);
            }
        }
        
        for(int k=ll;k<=rr;++k){
            int t=qs[k];
            if(((qx1[t]<=mid)&&(qx2[t]>=mid))||((qx1[t]>=mid)&&(qx2[t]<=mid)))continue;
            if(qx1[t]<=mid){
                tmpq[p1]=t;++p1;
            }
            if(qx1[t]>=mid){
                tmpq[p2]=t;--p2;
            }
        }
        for(int i=ll;i<=rr;++i)qs[i]=tmpq[i];
        DivCon(upx,upy,mid-1,downy,ll,p1-1);
        DivCon(mid+1,upy,downx,downy,p2+1,rr);
    }
}


int main(){
//    freopen("tourist.in","r",stdin);
//    freopen("tourist.out","w",stdout);
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;++i){
        for(int j=1;j<m;++j){
            scanf("%d",&rightdist[P(i,j)]);
        }
    }
    for(int i=1;i<=n-1;++i){
        for(int j=1;j<=m;++j){
            scanf("%d",&downdist[P(i,j)]);
        }
    }
    
    scanf("%d",&TT);
    for(int i=1;i<=TT;++i){
        scanf("%d%d%d%d",&qx1[i],&qy1[i],&qx2[i],&qy2[i]);
    }
    for(int i=1;i<=TT;++i)ans[i]=0x7fffffff;
    for(int i=1;i<=TT;++i)qs[i]=i;
    
    DivCon(1,1,n,m,1,TT);
    
    
    for(int i=1;i<=TT;++i)printf("%d\n",ans[i]);
    return 0;
}