Luogu3350 ZJOI2016 旅行者 最短路、分治

传送门

题意：给出一个$N \times M$的网格图，边有边权，$Q$组询问，每组询问$(x_1,y_1)$到$(x_2,y_2)$的最短路。$N \times M \leq 2 \times 10^4 , Q \leq 10^5$

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BZOJ原题竟然没有数据范围

矩形的多组询问问题考虑分治。考虑计算矩形$(x_1,y_1,x_2,y_2)$的询问，我们将较长边沿着中线劈成两半，在这些询问里面就只可能存在两种情况：①询问的两个端点在中线两边，那么路径就一定会经过中线上一点；②询问的两个端点在中线的同一边，那么有可能不会经过中线，也有可能经过中线。所以我们对中线上所有的点跑一边整个矩形的最短路，每一次跑完最短路就更新所有的询问，然后再递归进入两侧矩形回答②类型的询问。时间复杂度似乎是$O(NM \times \sqrt{NM} \times log(NM))$

在$Luogu$题解上学到的一个加速方法：从一个点的最短路转移到新的点跑最短路的时候，可以不将最短路数组赋值为极大值，而是赋值为这一个点经过上一个计算的点到达所有目标点的答案。

（然而还是在Luogu不开O2的情况下TLE）

 // luogu-judger-enable-o2
 //This code is written by Itst
 #include<bits/stdc++.h>
 #define pos(i,j) ((i-1)*M+j)
 using namespace std;

 inline int read(){
     int a = ;
     bool f = ;
     char c = getchar();
     while(c != EOF && !isdigit(c)){
         if(c == '-')
             f = ;
         c = getchar();
     }
     while(c != EOF && isdigit(c)){
         a = (a << ) + (a << ) + (c ^ '');
         c = getchar();
     }
     return f ? -a : a;
 }

 const int MAXN = ;
 int cntEd , N , M , Q , minDis[MAXN] , ans[MAXN] , head[MAXN];
 struct Edge{
     int end , upEd , w;
 }Ed[MAXN << ];
 struct query{
     int sx , sy , ex , ey , ind;
 }now[MAXN] , pot[MAXN];
 priority_queue < pair < int , int > > q;

 inline void addEd(int a , int b , int c){
     Ed[++cntEd].end = b;
     Ed[cntEd].w = c;
     Ed[cntEd].upEd = head[a];
     head[a] = cntEd;
 }

 void Dijk(int bx , int by , int lx , int ly , int rx , int ry , int l){
     int temp = minDis[pos(bx , by)];
     for(int i = lx ; i <= rx ; i++)
         for(int j = ly ; j <= ry ; j++)
             minDis[pos(i , j)] = l ==  ? 0x3f3f3f3f : minDis[pos(i , j)] + temp;
     minDis[pos(bx , by)] = ;
     q.push(make_pair( , pos(bx , by)));
     while(!q.empty()){
         pair < int , int > t = q.top();
         q.pop();
         if(minDis[t.second] != -t.first)
             continue;
         for(int i = head[t.second] ; i ; i = Ed[i].upEd)
             if(minDis[Ed[i].end] > minDis[t.second] + Ed[i].w){
                 minDis[Ed[i].end] = minDis[t.second] + Ed[i].w;
                 q.push(make_pair(-minDis[Ed[i].end] , Ed[i].end));
             }
     }
 }

 void solve(int lx , int ly , int rx , int ry , int ql , int qr){
     if(ql > qr)
         return;
     if(rx - lx > ry - ly){
         int mid = lx + rx >> ;
         for(int i = ly ; i <= ry ; i++){
             Dijk(mid , i , lx , ly , rx , ry , i - ly);
             for(int j = ql ; j <= qr ; j++)
                 ans[now[j].ind] = min(ans[now[j].ind] , minDis[pos(now[j].sx , now[j].sy)] + minDis[pos(now[j].ex , now[j].ey)]);
         }
         int p1 = ql , p2 = qr;
         for(int i = ql ; i <= qr ; i++)
             if(now[i].sx < mid && now[i].ex < mid)
                 pot[p1++] = now[i];
             else
                 if(now[i].sx > mid && now[i].ex > mid)
                     pot[p2--] = now[i];
         memcpy(now + ql , pot + ql , sizeof(query) * (qr - ql + ));
         solve(lx , ly , mid , ry , ql , p1 - );
         solve(mid +  , ly , rx , ry , p2 +  , qr);
     }
     else{
         int mid = ly + ry >> ;
         for(int i = lx ; i <= rx ; i++){
             Dijk(i , mid , lx , ly , rx , ry , i - lx);
             for(int j = ql ; j <= qr ; j++)
                 ans[now[j].ind] = min(ans[now[j].ind] , minDis[pos(now[j].sx , now[j].sy)] + minDis[pos(now[j].ex , now[j].ey)]);
         }
         int p1 = ql , p2 = qr;
         for(int i = ql ; i <= qr ; i++)
             if(now[i].sy < mid && now[i].ey < mid)
                 pot[p1++] = now[i];
             else
                 if(now[i].sy > mid && now[i].ey > mid)
                     pot[p2--] = now[i];
         memcpy(now + ql , pot + ql , sizeof(query) * (qr - ql + ));
         solve(lx , ly , rx , mid , ql , p1 - );
         solve(lx , mid +  , rx , ry , p2 +  , qr);
     }
 }

 int main(){
 #ifdef LG
     freopen("3350.in" , "r" , stdin);
     freopen("3350.out" , "w" , stdout);
 #endif
     memset(ans , 0x3f , sizeof(ans));
     N = read();
     M = read();
     for(int i =  ; i <= N ; i++)
         for(int j =  ; j < M ; j++){
             int t = read();
             addEd(pos(i , j) , pos(i , j + ) , t);
             addEd(pos(i , j + ) , pos(i , j) , t);
         }
     for(int i =  ; i < N ; i++)
         for(int j =  ; j <= M ; j++){
             int t = read();
             addEd(pos(i , j) , pos(i +  , j) , t);
             addEd(pos(i +  , j) , pos(i , j) , t);
         }
     Q = read();
     for(int i =  ; i <= Q ; i++){
         now[i].sx = read();
         now[i].sy = read();
         now[i].ex = read();
         now[i].ey = read();
         now[i].ind = i;
     }
     solve( ,  , N , M ,  , Q);
     for(int i =  ; i <= Q ; i++)
         printf("%d\n" , ans[i]);
     return ;
 }