CF 1422D Returning Home

题目描述

https://codeforces.com/contest/1422/problem/D

给一张网格图,图上有 $n$ 个特殊点,从任何一个与特殊点同行或者同列的点可以花费 $0$ 的时间到达特殊点,除此之外,每个点只能到周围四个点,花费 $1$ 的时间,给定起点和终点,询问到终点的最短时间

Solution

首先行和列离散化

对于每行和每列新建一个点

每个特殊点 $(x_i,y_i)$ 连 $x_i$ 和 $y_i$,双向边,权值为 $0$

每个特殊点 $(x_i,y_i)$ 连终点 $t$,权值为 $|x_i-tx|+|y_i-ty|$

起点 $(sx,sy)$ 连 $sx$ 和 $sy$,权值为 $0$

$x_i$ 连 $x_{i+1}$,权值为 $x_{i+1}-x_i$,双向边

$y_i$ 连 $y_{i+1}$,权值为 $y_{i+1}-y_i$,双向边

跑最短路即可

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define maxn 100010
#define Maxn 300010
#define cn const node
#define cQ const Queue
#define ll long long
#define INF 1000000000000000000ll
using namespace std;

int m, n;

int sx, sy, tx, ty;

struct node {
int x, y;

node() {}
node(int _x, int _y) { x = _x; y = _y; }
} a[maxn];

int X[maxn], Y[maxn], cnt1, cnt2;
void init_hash() {
for (int i = 1; i <= n; ++i) X[i] = a[i].x, Y[i] = a[i].y;
sort(X + 1, X + n + 1); cnt1 = unique(X + 1, X + n + 1) - X - 1;
sort(Y + 1, Y + n + 1); cnt2 = unique(Y + 1, Y + n + 1) - Y - 1;
for (int i = 1; i <= n; ++i) {
a[i].x = lower_bound(X + 1, X + cnt1 + 1, a[i].x) - X;
a[i].y = lower_bound(Y + 1, Y + cnt2 + 1, a[i].y) - Y;
}
}

struct Edge {
int to, next, w;
} e[8 * maxn]; int c1, head[Maxn];
inline void add_edge(int u, int v, int w) {
e[c1].to = v; e[c1].w = w;
e[c1].next = head[u]; head[u] = c1++;
}

inline void Add_edge(int u, int v, int w) {
add_edge(u, v, w); add_edge(v, u, w);
}

struct Queue {
int k; ll v;

Queue() {}
Queue(int _k, ll _v) { k = _k; v = _v; }

friend bool operator < (cQ &u, cQ &v) { return u.v > v.v; }
}; priority_queue<Queue> Q; bool vis[Maxn]; ll dis[Maxn]; int s, t;
void dijkstra() {
Q.push(Queue(s, 0));
for (int i = 1; i <= cnt1 + cnt2 + n; ++i) dis[i] = INF; dis[s] = 0;
while (!Q.empty()) {
int u = Q.top().k; Q.pop();
if (vis[u]) continue; vis[u] = 1;
for (int i = head[u]; ~i; i = e[i].next) {
int v = e[i].to, w = e[i].w;
if (dis[v] > dis[u] + w) {
dis[v] = dis[u] + w;
Q.push(Queue(v, dis[v]));
}
}
}
}

int main() { memset(head, -1, sizeof head);
cin >> m >> n >> sx >> sy >> tx >> ty;
for (int i = 1; i <= n; ++i) cin >> a[i].x >> a[i].y;
a[++n] = node(sx, sy); init_hash(); s = cnt1 + cnt2 + n;
for (int i = 1; i <= n; ++i) {
int x = a[i].x, y = a[i].y, id = i + cnt1 + cnt2;
Add_edge(id, x, 0); Add_edge(id, cnt1 + y, 0);
}
for (int i = 1; i < cnt1; ++i) Add_edge(i, i + 1, X[i + 1] - X[i]);
for (int i = 1; i < cnt2; ++i) Add_edge(cnt1 + i, cnt1 + i + 1, Y[i + 1] - Y[i]);
dijkstra(); ll ans = INF;
for (int i = 1; i <= n; ++i)
ans = min(ans, dis[i + cnt1 + cnt2] + abs(X[a[i].x] - tx) + abs(Y[a[i].y] - ty));
cout << ans << endl;
return 0;
}