[国家集训队] 飞飞侠 题解

前言

题目链接：洛谷。

题意分析

显然需要三次单源最短路，但不能朴素建图。曼哈顿转成切比雪夫，然后 KD-T 或二维线段树优化建图即可。

最多 \(\mathcal{O}(n^2\times n)\) 条边，时间复杂度 \(\mathcal{O}(n^3\log n)\)。

代码

#include <cstdio>
#include <iostream>
#include <vector>
#include <queue>
using namespace std;using ll = long long;template <typename T>
using minHeap = priority_queue<T, vector<T>, greater<T>>;const int N = 150 + 10;
const ll inf = 0x3f3f3f3f3f3f3f3f;int n, m;
int b[N][N], a[N][N];inline int id(int x, int y) {return (x - 1) * m + y;
}struct P {int x, y;void R() {scanf("%d%d", &x, &y);}int operator()() const {return id(x, y);}
} X, Y, Z;namespace sub1 {bool check() {if (n > 100 || m > 100)return false;for (int i = 1; i <= n; ++i)for (int j = 1; j <= m; ++j)if (b[i][j] > 20)return false;return true;
}const int N = 1e4 + 10;vector<pair<int, int>> e[N];ll dis[N];
ll solve(P x, P y1, P y2) {for (int i = 1; i <= n * m; ++i) {dis[i] = inf;}minHeap<pair<ll, int>> Q;Q.emplace(dis[x()] = 0, x());while (!Q.empty()) {ll ndis = Q.top().first;int u = Q.top().second;Q.pop();if (dis[u] < ndis) continue;for (auto E : e[u]) {int v = E.first;int w = E.second;if (dis[v] > dis[u] + w)Q.emplace(dis[v] = dis[u] + w, v);}}if (dis[y1()] == inf || dis[y2()] == inf)return inf;return dis[y1()] + dis[y2()];
}void solve() {for (int i = 1; i <= n; ++i)for (int j = 1; j <= m; ++j) {for (int ti = max(1, i - b[i][j]); ti <= min(n, i + b[i][j]); ++ti)for (int tj = max(1, j - b[i][j]); tj <= min(m, j + b[i][j]); ++tj)if (ti != i || tj != j) {if (abs(i - ti) + abs(j - tj) > b[i][j])continue;// printf("(%d, %d) ---(%d)---> (%d, %d)\n", i, j, a[i][j], ti, tj);e[id(ti, tj)].emplace_back(id(i, j), a[i][j]);}}ll ans = inf, res;char chr = 0;if ((res = solve(X, Y, Z)) < ans) {ans = res;chr = 'X';}if ((res = solve(Y, X, Z)) < ans) {ans = res;chr = 'Y';}if ((res = solve(Z, X, Y)) < ans) {ans = res;chr = 'Z';}if (chr == 0) {puts("NO");} else {printf("%c\n%lld\n", chr, ans);}
}}namespace yzh {// KD-Tconst int M = 111900 * 3; // N * N * N * 4;
const int NM = M + N * N;int rt, son[M][2], tot;vector<pair<int, int>> e[NM];void build(int &u, int xl, int xr, int yl, int yr) {if (xl == xr && yl == yr) {int x = xl, y = yl;if ((x + y) & 1) return;int a = (x + y) / 2;int b = (y - x) / 2;if (a < 1 || a > n || b < 1 || b > m) return;u = ++tot;e[id(a, b)].emplace_back(n * m + u, 0);return;}u = ++tot;if (xr - xl > yr - yl) {int mid = (xl + xr) >> 1;build(son[u][0], xl, mid, yl, yr);build(son[u][1], mid + 1, xr, yl, yr);} else {int mid = (yl + yr) >> 1;build(son[u][0], xl, xr, yl, mid);build(son[u][1], xl, xr, mid + 1, yr);}if (son[u][0])e[n * m + son[u][0]].emplace_back(n * m + u, 0);if (son[u][1])e[n * m + son[u][1]].emplace_back(n * m + u, 0);
}void adde(int u, int txl, int txr, int tyl, int tyr, int xl, int xr, int yl, int yr, int x, int w) {if (!u) return;if (xl <= txl && txr <= xr && yl <= tyl && tyr <= yr) {e[n * m + u].emplace_back(x, w);return;}if (txr - txl > tyr - tyl) {int mid = (txl + txr) >> 1;if (xl <= mid) adde(son[u][0], txl, mid, tyl, tyr, xl, xr, yl, yr, x, w);if (xr > mid) adde(son[u][1], mid + 1, txr, tyl, tyr, xl, xr, yl, yr, x, w);} else {int mid = (tyl + tyr) >> 1;if (yl <= mid) adde(son[u][0], txl, txr, tyl, mid, xl, xr, yl, yr, x, w);if (yr > mid) adde(son[u][1], txl, txr, mid + 1, tyr, xl, xr, yl, yr, x, w);}
}ll dis[NM];
ll solve(P x, P y1, P y2) {for (int i = 1; i <= n * m + tot; ++i) {dis[i] = inf;}minHeap<pair<ll, int>> Q;Q.emplace(dis[x()] = 0, x());while (!Q.empty()) {ll ndis = Q.top().first;int u = Q.top().second;Q.pop();if (dis[u] < ndis) continue;for (auto E : e[u]) {int v = E.first;int w = E.second;if (dis[v] > dis[u] + w)Q.emplace(dis[v] = dis[u] + w, v);}}if (dis[y1()] == inf || dis[y2()] == inf)return inf;return dis[y1()] + dis[y2()];
}void adde(int xl, int xr, int yl, int yr, int x, int w) {xl = max(xl, 1 - m);xr = min(xr, n - 1);yl = max(yl, 2);yr = min(yr, n + m);if (xl > xr || yl > yr) return;adde(rt, 1 - m, n - 1, 1 + 1, n + m, xl, xr, yl, yr, x, w);
}void solve() {build(rt, 1 - m, n - 1, 1 + 1, n + m);fprintf(stderr, "tot = %d\n", tot);for (int i = 1; i <= n; ++i)for (int j = 1; j <= m; ++j) {if (b[i][j]) {adde(i - j - b[i][j], i - j + b[i][j], i + j - b[i][j], i + j + b[i][j], id(i, j), a[i][j]);}}ll ans = inf, res;char chr = 0;if ((res = solve(X, Y, Z)) < ans) {ans = res;chr = 'X';}fprintf(stderr, "res = %lld\n", res);if ((res = solve(Y, X, Z)) < ans) {ans = res;chr = 'Y';}fprintf(stderr, "res = %lld\n", res);if ((res = solve(Z, X, Y)) < ans) {ans = res;chr = 'Z';}fprintf(stderr, "res = %lld\n", res);if (chr == 0) {puts("NO");} else {printf("%c\n%lld\n", chr, ans);}
}}int main() {scanf("%d%d", &n, &m);for (int i = 1; i <= n; ++i) {for (int j = 1; j <= m; ++j) {scanf("%d", &b[i][j]);}}for (int i = 1; i <= n; ++i) {for (int j = 1; j <= m; ++j) {scanf("%d", &a[i][j]);}}X.R(), Y.R(), Z.R();
#ifndef XuYuemingif (sub1::check()) return sub1::solve(), 0;
#endifyzh::solve();return 0;
}// g++ -O2 -Wall -Wextra -std=c++14 -DXuYueming 3.cpp -o 3