题目
Sol
离线+分治+(djikstra)
每次把网格图分成两部分
如果起点和终点分隔两边，那么它一定会经过中轴线
枚举中轴线上的点跑(dijkstra)
然后处理询问，递归处理起点终点在一起的询问
# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(1e5 + 5);
const int INF(1e9);
IL int Input(){
RG int x = 0, z = 1; RG char c = getchar();
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}
int n, m, q, dis[_], ans[_];
struct Query{
int x1, y1, x2, y2, s, t, id;
} qry[_], tmp1[_], tmp2[_];
int vis[_], first[_], cnt;
struct Edge{
int to, next, w;
} edge[_ << 1];
struct Data{
int u, w;
IL bool operator <(RG Data B) const{
return w > B.w;
}
};
priority_queue <Data> Q;
IL void Add(RG int u, RG int v, RG int w){
edge[cnt] = (Edge){v, first[u], w}, first[u] = cnt++;
}
# define ID(x, y) (x - 1) * m + y
IL void Dijkstra(RG int l1, RG int r1, RG int l2, RG int r2, RG int S){
for(RG int i = l1; i <= l2; ++i)
for(RG int j = r1; j <= r2; ++j){
RG int id = ID(i, j);
dis[id] = INF, vis[id] = 0;
}
dis[S] = 0, Q.push((Data){S, 0});
while(!Q.empty()){
RG Data s = Q.top(); Q.pop();
if(vis[s.u]) continue;
vis[s.u] = 1;
for(RG int e = first[s.u]; e != -1; e = edge[e].next){
RG int v = edge[e].to, w = edge[e].w;
if(vis[v]) continue;
if(dis[s.u] + w < dis[v]){
dis[v] = dis[s.u] + w;
Q.push((Data){v, dis[v]});
}
}
}
}
IL void Solve(RG int l1, RG int r1, RG int l2, RG int r2, RG int l, RG int r){
if(l1 == l2 && r1 == r2){
for(RG int i = l; i <= r; ++i) ans[qry[i].id] = 0;
return;
}
if(l2 - l1 > r2 - r1){
RG int mid = (l1 + l2) >> 1, cnt1 = 0, cnt2 = 0;
for(RG int i = r1; i <= r2; ++i){
RG int p = ID(mid, i);
Dijkstra(l1, r1, l2, r2, p);
for(RG int j = l; j <= r; ++j)
ans[qry[j].id] = min(ans[qry[j].id], dis[qry[j].s] + dis[qry[j].t]);
}
for(RG int i = l; i <= r; ++i)
if(qry[i].x1 <= mid && qry[i].x2 <= mid) tmp1[++cnt1] = qry[i];
else if(qry[i].x1 > mid && qry[i].x2 > mid) tmp2[++cnt2] = qry[i];
for(RG int i = 1; i <= cnt1; ++i) qry[l + i - 1] = tmp1[i];
for(RG int i = 1; i <= cnt2; ++i) qry[cnt1 + l - 1 + i] = tmp2[i];
Solve(l1, r1, mid, r2, l, cnt1 + l - 1), Solve(mid + 1, r1, l2, r2, cnt1 + l, l + cnt2 + cnt1 - 1);
}
else{
RG int mid = (r1 + r2) >> 1, cnt1 = 0, cnt2 = 0;
for(RG int i = l1; i <= l2; ++i){
RG int p = ID(i, mid);
Dijkstra(l1, r1, l2, r2, p);
for(RG int j = l; j <= r; ++j)
ans[qry[j].id] = min(ans[qry[j].id], dis[qry[j].s] + dis[qry[j].t]);
}
for(RG int i = l; i <= r; ++i)
if(qry[i].y1 <= mid && qry[i].y2 <= mid) tmp1[++cnt1] = qry[i];
else if(qry[i].y1 > mid && qry[i].y2 > mid) tmp2[++cnt2] = qry[i];
for(RG int i = 1; i <= cnt1; ++i) qry[l + i - 1] = tmp1[i];
for(RG int i = 1; i <= cnt2; ++i) qry[cnt1 + l - 1 + i] = tmp2[i];
Solve(l1, r1, l2, mid, l, cnt1 + l - 1), Solve(l1, mid + 1, l2, r2, cnt1 + l, l + cnt1 + cnt2 - 1);
}
}
int main(RG int argc, RG char *argv[]){
n = Input(), m = Input();
for(RG int i = 1; i <= n; ++i)
for(RG int j = 1; j <= m; ++j){
RG int id = ID(i, j);
vis[id] = 1, first[id] = -1;
}
for(RG int i = 1; i <= n; ++i)
for(RG int j = 1; j < m; ++j){
RG int u = ID(i, j), v = ID(i, j + 1), w = Input();
Add(u, v, w), Add(v, u, w);
}
for(RG int i = 1; i < n; ++i)
for(RG int j = 1; j <= m; ++j){
RG int u = ID(i, j), v = ID(i + 1, j), w = Input();
Add(u, v, w), Add(v, u, w);
}
q = Input();
for(RG int i = 1; i <= q; ++i){
RG int x1 = Input(), y1 = Input(), s = ID(x1, y1), x2 = Input(), y2 = Input(), t = ID(x2, y2);
qry[i] = (Query){x1, y1, x2, y2, s, t, i}, ans[i] = INF;
}
Solve(1, 1, n, m, 1, q);
for(RG int i = 1; i <= q; ++i) printf("%d
", ans[i]);
return 0;
}