4456: [Zjoi2016]旅行者

4456: [Zjoi2016]旅行者

https://www.lydsy.com/JudgeOnline/problem.php?id=4456

分析：

　　每次对当前矩阵按长边化一条分治线，然后在对分治线上的点跑最短路，然后可以处理处过分治线的询问。对于不过分治线的，递归处理。

　　先写的dijkstra+堆优化，在开O2的情况下可以过，不开O2过不了，卡常~，还是过不了。然后在UOJ的排行榜里（而且UOJ是有大样例的！），看到了两个优化，加上就可以了。

　　优化：1、代码41行，用上次的遍历结果初始化。2、将dijkstra改成spfa+SLF优化。

代码：

 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<cmath>
 #include<iostream>
 using namespace std;
 typedef long long LL;

 char buf[],*_p1 = buf,*_p2 = buf;
 #define nc() (_p1==_p2&&(_p2=(_p1=buf)+fread(buf,1,100000,stdin),_p1==_p2) ? EOF :*_p1++)
 inline int read() {
     int x=,f=;char ch=nc();for(;!isdigit(ch);ch=nc())if(ch=='-')f=-;
     for (;isdigit(ch);ch=nc())x=x*+ch-'';return x*f;
 } 

 #define getid(a,b) ((a - 1) * m + b)
 #define pa pair<int,int>
 #define mp(a,b) make_pair(a,b)

 const int INF = 1e9;
 const int N = ;

 struct Que{
     int a1,b1,a2,b2,id;
 }A[N], B[N];
 int dis[N], ans[N], head[N], nxt[N], to[N], len[N];
 int n, m, En;
 //priority_queue< pa, vector< pa >, greater< pa > > q; // 这才是小根堆！！！
 int q[];
 bool vis[N];

 inline void getxy(int id,int &x,int &y) {
     x = (id - ) / m + ;
     y = (id - ) % m + ;
 }
 void dijkstra(int s,int a1,int a2,int b1,int b2,int h) {
     int x, y, d = dis[s];
     for (int i=a1; i<=a2; ++i)
         for (int j=b1; j<=b2; ++j) {
             x = getid(i, j);
             dis[x] = h ? dis[x] + d : INF;
             vis[x] = false;
         }
     dis[s] = ;
 /*    q.push(mp(dis[s], s));
     while (!q.empty()) {
         pa now = q.top(); q.pop();
         int u = now.second;
         if (vis[u]) continue;
         vis[u] = true;
         for (int i=head[u]; i; i=nxt[i]) {
             int v = to[i];
             getxy(v, x, y);
             if (x >= a1 && x <= a2 && y >= b1 && y <= b2 && dis[v] > dis[u] + len[i]) {
                 dis[v] = dis[u] + len[i];
                 q.push(mp(dis[v], v));
             }
         }
     }*/
     int L = , R = L - ;
     q[++R] = s;vis[s] = true;
     while (L <= R) {
         int u = q[L ++];vis[u] = false;
         for (int i=head[u]; i; i=nxt[i]) {
             int v = to[i];
             getxy(v, x, y);
             if (x >= a1 && x <= a2 && y >= b1 && y <= b2 && dis[v] > dis[u] + len[i]) {
                 dis[v] = dis[u] + len[i];
                 if (!vis[v]) {
                     if (dis[v] <= dis[q[L]]) q[--L] = v;
                     else q[++R] = v;
                     vis[v] = true;
                 }
             }
         }
     }
 }
 void solve(int a1,int a2,int b1,int b2,int L,int R) {
     if (L > R) return ;
     int i, j, k;
     if (a2 - a1 > b2 - b1) {
         int mid = (a2 + a1) / ; // a2 + a1
         for (i=b1; i<=b2; ++i) {
             dijkstra(getid(mid, i), a1, a2, b1, b2, i-b1);
             for (j=L; j<=R; ++j)
                 ans[A[j].id] = min(ans[A[j].id], dis[getid(A[j].a1, A[j].b1)] + dis[getid(A[j].a2, A[j].b2)]);
         }
         i = L - , j = R + ;
         for (k=L; k<=R; ++k) {
             if (A[k].a1 < mid && A[k].a2 < mid) B[++i] = A[k];
             if (A[k].a1 > mid && A[k].a2 > mid) B[--j] = A[k];
         }
         for (k=L; k<=i; ++k) A[k] = B[k]; solve(a1, mid-, b1, b2, L, i);
         for (k=j; k<=R; ++k) A[k] = B[k]; solve(mid+, a2, b1, b2, j, R);
     }
     else {
         int mid = (b2 + b1) / ;
         for (i=a1; i<=a2; ++i) {
             dijkstra(getid(i, mid), a1, a2, b1, b2, i-a1);
             for (j=L; j<=R; ++j)
                 ans[A[j].id] = min(ans[A[j].id], dis[getid(A[j].a1, A[j].b1)] + dis[getid(A[j].a2, A[j].b2)]);
         }
         i = L - , j = R + ;
         for (k=L; k<=R; ++k) {
             if (A[k].b1 < mid && A[k].b2 < mid) B[++i] = A[k];
             if (A[k].b1 > mid && A[k].b2 > mid) B[--j] = A[k];
         }
         for (k=L; k<=i; ++k) A[k] = B[k]; solve(a1, a2, b1, mid-, L, i);
         for (k=j; k<=R; ++k) A[k] = B[k]; solve(a1, a2, mid+, b2, j, R);
     }
 }
 inline void add_edge(int u,int v,int w) {
     ++En; to[En] = v; len[En] = w; nxt[En] = head[u]; head[u] = En;
     ++En; to[En] = u; len[En] = w; nxt[En] = head[v]; head[v] = En;
 }
 int main() {
     n = read(), m = read();
     for (int i=; i<=n; ++i) {
         for (int j=; j<m; ++j) {
             int a = read();
             add_edge(getid(i, j), getid(i, j + ), a);
         }
     }
     for (int i=; i<n; ++i) {
         for (int j=; j<=m; ++j) {
             int a = read();
             add_edge(getid(i, j), getid(i + , j), a);
         }
     }
     int Case = read();
     for (int a,b,i=; i<=Case; ++i) {
         A[i].a1 = read(), A[i].b1 = read();
         A[i].a2 = read(), A[i].b2 = read();
         A[i].id = i; ans[i] = INF;
     }
     solve(, n, , m, , Case);
     for (int i=; i<=Case; ++i) printf("%d\n",ans[i]);
     return ;
 }