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 ;
}