P3350 [ZJOI2016]旅行者

題目描述：

Link

題目分析：

其實就是求 網格圖多源多匯最短路 直接做看起來不太可做。

但是。重要的事情說三遍。

網格圖先想分治，網格圖先想分治，網格圖先想分治

類似於整體二分，首先肯定是要把所有的操作都離線下來的。然後我們考慮對於這個網格圖，類似於 KDT 的那種建圖的分治方式去做。

首先劃分這個矩形的時候，肯定要先劃分較長的一邊，那麼此時，我們就相當於在中間畫了一條豎線，此時分為兩種情況。

1. 詢問不越過這條線，那直接往下進行分治即可
2. 詢問跨過這條線，那麼直接列舉中間那條邊上的每一個點，更新詢問即可。

複雜度貌似是 \(S\sqrt{S}\log S\)

複雜度證明

Code：

//editor : DRYAYST
//Wo shi ge da SHA BI
//under O2
#include<bits/stdc++.h>
#define g() getchar()
#define il inline
#define ull unsigned long long
#define eps 1e-10
#define ll long long
#define pa pair<int, int>
#define for_1(i, n) for(int i = 1; i <= (n); ++i)
#define for_0(i, n) for(int i = 0; i < (n); ++i)
#define for_xy(i, x, y) for(int i = (x); i <= (y); ++i)
#define for_yx(i, y, x) for(int i = (y); i >= (x); --i)
#define for_edge(i, x) for(int i = head[x]; i; i = nxt[i])
#define int long long
#define DB double
#define ls (p<<1)
#define rs (p<<1|1)
#define m_p make_pair
#define fi first
#define se second
using namespace std;
const int N = 1e6 + 10, INF = 0x7f7f7f7f, mod = 1e9 + 7;
il int qpow(int x, int k) {int ans = 1; while(k) {if(k & 1) ans = ans * x % mod; x = x * x % mod; k >>= 1; } return ans; }
il int Add(int x, int y) {return (x += y) %= mod;}
il int Del(int x, int y) {return (x = x - y + mod) % mod;}
il int Mul(int x, int y) {return x * y % mod;}
il int inv(int x) {return qpow(x, mod - 2); }
il int re() {
    int x = 0, p = 1;
    char ch = getchar();
    while(ch > '9' || ch < '0') {if(ch == '-') p = -1; ch = getchar();}
    while(ch <= '9' and ch >= '0') {x = (x << 3) + (x << 1) + (ch ^ 48); ch = getchar();}
    return x * p;
}
struct Query{ int x1 , x2, y1 , y2, id;} Q[N], Tp[N];
struct node{ int x, y, w; friend bool operator < (node x , node y){ return x.w > y.w; } };
priority_queue<node> q;
int dx[] = {0  , -1 , 0 , 1}, dy[] = {-1 , 0  , 1 , 0};
int ans[N], dis[N], w[N][4], a[N], b[N];
int n, m;
il int id(int x , int y) { return (x - 1) * m + y; }
il void dij(int l , int r , int x , int y , int x1 , int y1 , int x2 , int y2) {
    node now = node{x , y , 0}; q.push(now);
    for(int i = x1; i <= x2; i++) { for(int j = y1; j <= y2; j++) { dis[id(i, j)] = INF; } } dis[id(x , y)] = 0;
    while(!q.empty()){
        now = q.top(), q.pop(); x = now.x, y = now.y; int pos = id(x, y); if(dis[pos] != now.w) continue;
        for(int i = 0; i < 4; i++) {
            if(x + dx[i] < x1 || x + dx[i] > x2 || y + dy[i] < y1 || y + dy[i] > y2) continue;
            int to = id(x + dx[i], y + dy[i]);
            if(dis[to] > dis[pos] + w[pos][i]){ dis[to] = dis[pos] + w[pos][i]; now = node{x + dx[i] , y + dy[i] , dis[to]}; q.push(now); }
		}
	}
    for(int i = l ; i <= r ; i++){ ans[Q[i].id] = min(ans[Q[i].id] , dis[id(Q[i].x1, Q[i].y1)] + dis[id(Q[i].x2, Q[i].y2)]); }
}
void solve(int l , int r , int x1 , int x2 , int y1 , int y2){
    if(x2 - x1 >= y2 - y1){
        int mid = (x1 + x2) >> 1; for(int i = y1; i <= y2; i++) dij(l, r, mid, i, x1, y1, x2, y2);
        int L = l - 1, R = r + 1; for(int i = l; i <= r; i++) Tp[i] = Q[i];
        for(int i = l; i <= r; i++) { if(Tp[i].x1 < mid && Tp[i].x2 < mid) Q[++L] = Tp[i]; if(Tp[i].x1 > mid && Tp[i].x2 > mid) Q[--R] = Tp[i]; }
		if(l <= L) solve(l, L, x1, mid - 1, y1, y2); if(R <= r) solve(R, r, mid + 1, x2, y1, y2);
    }
	else{
        int mid = (y1 + y2) >> 1; for(int i = x1; i <= x2; i++) dij(l, r, i, mid, x1, y1, x2, y2);
        int L = l - 1, R = r + 1; for(int i = l; i <= r; i++) Tp[i] = Q[i];
        for(int i = l; i <= r; i++) { if(Tp[i].y1 < mid && Tp[i].y2 < mid) Q[++L] = Tp[i]; if(Tp[i].y1 > mid && Tp[i].y2 > mid) Q[--R] = Tp[i];  }
        if(l <= L) solve(l, L, x1, x2, y1, mid - 1); if(R <= r) solve(R, r, x1, x2, mid + 1, y2);
	}
}
signed main() {
    n = re(), m = re(); 
    for_1(i, n) for_1(j, m-1) a[id(i,j)] = re();
    for_1(i, n-1) for_1(j, m) b[id(i,j)] = re(); 
    for_1(i,n) for_1(j, m) { 
        int x = id(i , j);
        for(int k = 0; k < 4; k++) {
            if(i + dx[k] < 1 || i + dx[k] > n || j + dy[k] < 1 || j + dy[k] > m) continue;
            if(k == 0) w[x][k] = a[id(i , j - 1)]; if(k == 1) w[x][k] = b[id(i - 1 , j)];
            if(k == 2) w[x][k] = a[id(i , j)]; if(k == 3) w[x][k] = b[id(i , j)];
        }
	}
    int qnum = re();  for(int i = 1; i <= qnum; i++) { Q[i].x1 = re(), Q[i].y1 = re(), Q[i].x2 = re(), Q[i].y2 = re(); Q[i].id = i;  }
	memset(ans, 0x3f, sizeof(ans));  solve(1, qnum, 1, n, 1, m);
    for_1(i, qnum) {printf("%lld\n", ans[i]); }
}