[QOJ888] Travel around China 题解

[QOJ888] Travel around China 题解

Petrozavodsk Winter 2021. Day 4. PKU Contest (Common Contest 1)

考虑 \(n = 2\)，猫树分治，考虑统计所有经过 \(mid\) 的区间，从 \(mid\) 开始跑最短路，处理出区间左右端点到 \(mid\) 列的最短路，最后要求的形如 \(\sum_{x\in Left, y\in Right} \min(d_{1,i,mid}+d_{1,mid,j}, d_{2, i, mid} + d_{2, mid, j})\)，分类讨论谁取 \(\min\)，就是一个二维偏序。

考虑 \(n = 3\)，同样的猫树分治，不过这里会多一种情况，就是 \((1, i)\) 可能绕一圈到达 \((3, i)\)，设 \(f_i\) 到达 \((3,i)\) 的最小代价，转移有：向左走、向右走、直接向下走三种情况，那么可以用 Dijkstra 求出 \(f_i\)，最后还是分类讨论 \(\min\) 取谁做二维偏序即可。

代码难度较高。

时间复杂度：\(O(n\log ^2n)\)。

#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#define int long long
#define x first
#define y second
using namespace std;
typedef unsigned long long ull;
typedef pair<int, int> PII;
const int N = 5e5 + 10, mod = 1e9 + 7, INF = 1e18;

int n, m, a[4][N], f[N], ans, d[4][4][N];
struct qwq {
    int d, r, c;
    bool operator < (const qwq &W) const {
    return d < W.d;
    }
    bool operator > (const qwq &W) const {
    return d > W.d;
    }
};
priority_queue<qwq, vector<qwq>, greater<qwq> > heap;
void trans(int o, int r, int c, int v) {
    if(d[o][r][c] > v)
    d[o][r][c] = v, heap.push({v, r, c});
}
void dijkstra(int o, int s, int l, int r) {
    for(int i = 1; i <= 3; i ++) for(int j = l; j <= r; j ++) d[o][i][j] = INF;
    heap.push({a[o][s], o, s});
    d[o][o][s] = a[o][s];
    while(heap.size()) {
    auto t = heap.top(); heap.pop();
    if(t.d != d[o][t.r][t.c]) continue;
    if(t.r > 1) trans(o, t.r - 1, t.c, t.d + a[t.r - 1][t.c]);
    if(t.r < 3) trans(o, t.r + 1, t.c, t.d + a[t.r + 1][t.c]);
    if(t.c < r) trans(o, t.r, t.c + 1, t.d + a[t.r][t.c + 1]);
    if(t.c > l) trans(o, t.r, t.c - 1, t.d + a[t.r][t.c - 1]);
    }
}
void chkmin(int &a, int b) {
    if(a > b) a = b;
}
int disc[N], tot;
int findl(int x) {
    return upper_bound(disc + 1, disc + tot + 1, x) - disc - 1;
}
int findl2(int x) {
    return lower_bound(disc + 1, disc + tot + 1, x) - disc - 1;
}
int tr[N], tr2[N];
void update(int u, int c, int d) {
    for(; u <= tot; u += u & -u)
    tr[u] += c, tr2[u] += d;
}
int query(int u) {
    int sum = 0; for(; u; u &= u - 1)
    sum += tr[u];
    return sum;
}
int query2(int u) {
    int sum = 0; for(; u; u &= u - 1)
    sum += tr2[u];
    return sum % mod;
}
int F(int i, int j, int x, int y) {
    return d[x][i][j] - d[y][i][j];
}
void solve(int l, int r) {
    if(l == r) {
    (ans += (a[1][l] + a[2][l] * 2 + a[3][l] + f[l]) % mod) %= mod;
    return ;
    }
    int mid = l + r >> 1;
    solve(l, mid), solve(mid + 1, r);
    for(int o = 1; o <= 3; o ++) dijkstra(o, mid, l, r);
    for(int o = 1; o <= 3; o ++) for(int i = l; i <= r; i ++) chkmin(d[o][1][i], d[o][3][i] + f[i] - a[3][i]), chkmin(d[o][3][i], d[o][1][i] + f[i] - a[1][i]);
    for(int o = 1; o <= 3; o ++) for(int rw = 1; rw <= 3; rw ++) for(int i = mid+ 1; i <= r; i ++) d[o][rw][i] -= a[o][mid];
    vector<qwq> t1, t2;
    tot = 0;
    for(int i = 1; i <= 3; i ++) for(int j = l; j <= r; j ++) 
    if(j <= mid) disc[++ tot] = F(i, j, 1, 3), t1.push_back({F(i, j, 1, 2), i, j});
    else t2.push_back({F(i, j, 2, 1), i, j});
    sort(t1.begin(), t1.end()), sort(t2.begin(), t2.end());
    sort(disc + 1, disc + tot + 1), tot = unique(disc + 1, disc + tot + 1) - disc - 1;
    for(int i = 0, j = 0; j < t2.size(); j ++) {
    while(i < t1.size() && t1[i].d <= t2[j].d) update(findl(F(t1[i].r, t1[i].c, 1, 3)), 1, d[1][t1[i].r][t1[i].c]), i ++;
    int pos = findl(F(t2[j].r, t2[j].c, 3, 1));
    (ans += (query(pos) * d[1][t2[j].r][t2[j].c] + query2(pos)) % mod);
    }
    for(int i = 1; i <= tot; i ++) tr[i] = tr2[i] = 0;
    
    t1.clear(), t2.clear();
    tot = 0;
    for(int i = 1; i <= 3; i ++) for(int j = l; j <= r; j ++) 
    if(j <= mid) disc[++ tot] = F(i, j, 2, 1), t1.push_back({F(i, j, 2, 3), i, j});
    else t2.push_back({F(i, j, 3, 2), i, j});
    sort(t1.begin(), t1.end()), sort(t2.begin(), t2.end());
    sort(disc + 1, disc + tot + 1), tot = unique(disc + 1, disc + tot + 1) - disc - 1;
    for(int i = 0, j = 0; j < t2.size(); j ++) {
    while(i < t1.size() && t1[i].d <= t2[j].d) update(findl(F(t1[i].r, t1[i].c, 2, 1)), 1, d[2][t1[i].r][t1[i].c]), i ++;
    int pos = findl2(F(t2[j].r, t2[j].c, 1, 2));
    (ans += (query(pos) * d[2][t2[j].r][t2[j].c] + query2(pos)) % mod);
    }
    for(int i = 1; i <= tot; i ++) tr[i] = tr2[i] = 0;

    t1.clear(), t2.clear();
    tot = 0;
    for(int i = 1; i <= 3; i ++) for(int j = l; j <= r; j ++) 
    if(j <= mid) disc[++ tot] = F(i, j, 3, 2), t1.push_back({F(i, j, 3, 1), i, j});
    else t2.push_back({F(i, j, 1, 3), i, j});
    sort(t1.begin(), t1.end()), sort(t2.begin(), t2.end());
    sort(disc + 1, disc + tot + 1), tot = unique(disc + 1, disc + tot + 1) - disc - 1;
    for(int i = 0, j = 0; j < t2.size(); j ++) {
    while(i < t1.size() && t1[i].d < t2[j].d) update(findl(F(t1[i].r, t1[i].c, 3, 2)), 1, d[3][t1[i].r][t1[i].c]), i ++;
    int pos = findl2(F(t2[j].r, t2[j].c, 2, 3));
    (ans += (query(pos) * d[3][t2[j].r][t2[j].c] + query2(pos)) % mod);
    }
    ans %= mod;
    for(int i = 1; i <= tot; i ++) tr[i] = tr2[i] = 0;
}

signed main() {
//    freopen("ex_data.in", "r", stdin);
    ios::sync_with_stdio(0), cin.tie(0);
    cin >> n >> m;
    for(int i = 1; i <= 3; i ++) for(int j = 1; j <= m; j ++) cin >> a[i][j];
    priority_queue<PII, vector<PII>, greater<PII> > q;
    for(int i = 1; i <= m; i ++) f[i] = a[1][i] + a[2][i] + a[3][i], q.push({f[i], i});
    while(q.size()) {
    auto t = q.top().y, v = q.top().x; q.pop();
    if(v != f[t]) continue;
//    cout << v << '\n';
    if(t > 1) {
    if(f[t] + a[1][t - 1] + a[3][t - 1] < f[t - 1])
    f[t - 1] = f[t] + a[1][t - 1] + a[3][t - 1], q.push({f[t - 1], t - 1});
    }
    if(t < m) {
    if(f[t] + a[1][t + 1] + a[3][t + 1] < f[t + 1]) 
    f[t + 1] = f[t] + a[1][t + 1] + a[3][t + 1], q.push({f[t + 1], t + 1});
    }
    } 
    solve(1, m);
    cout << ans * 2 % mod << '\n';
    
    return 0;
}