洛谷 P3831 - 做题记录

原题链接

  • 洛谷 P3831 [SHOI2012] 回家的路

题目信息

使用的算法

  • 图论
  • 最短路

难度

  • 洛谷:提高 +/ 省选 -
  • 本人:0.7

简化题意

给出一个 n×nn \times n 的地图,在这个地图上有 mm 个点可以使横向与纵向的直线(线段)相连, 经过一条线段所花费的时间为 22, 方向改变所花费的时间为 11,询问从某一点 (xfrom,yfrom)(x_{from}, y_{from}) 出发,到 (xto,yto)(x_{to}, y_{to}) 的距离最短是多少(无法到达则为 1-1)。

思想方法

  • 到达的最短时间我们可以很容易想到使用最短路进行求解
  • 可以将每个交汇点当作图中的一个节点进行存储
  • 每个节点之间的距离保存为他们之间的线段数量的两倍

代码实现

保存交汇节点

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class Station {
public:
unsigned int id, x, y;

Station(const unsigned int &id, const unsigned int &x,
const unsigned int &y) {
this->id = id;
this->x = x;
this->y = y;
}
};

存边

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class Edge {
public:
unsigned int from, to, len;

Edge(const unsigned int &from, const unsigned int &to,
const unsigned int &len) {
this->from = from;
this->to = to;
this->len = len;
}
};

存图

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unsigned int n, m;
vector<Station> stations;
vector<Edge> graph[MAX_M];

首先判断答案是否为 1-1

  • 当起点所在的坐标 (xfrom,yfrom)(x_{from}, y_{from}) 时, 如没有坐标为 (xfrom,y)(x_{from}, y)(x,yfrom)(x, y_{from}) 的交汇点时无法到达终点, 因为无法通过交汇点到达目标(除非目标的任一坐标与起点相同),目标同理
  • 即可想到使用 set 存储交汇点的 (x,y)(x, y) 坐标
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set<unsigned int> coordinateX;
set<unsigned int> coordinateY;
if ((!coordinateX.count(xFrom) and !coordinateY.count(yFrom)) or
(!coordinateX.count(xTo) and !coordinateY.count(yTo))) {
cout << -1 << endl;
exit(0);
}

接着进行建图

  • 先将起点与目标当作交汇点存入 stations
  • 遍历 stations,将 xx 坐标相同的节点互相连接起来(无向边), yy 坐标同理
  • 为了更好的遍历,我们需要先对 stations 进行排序
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inline bool compX(const Station &x, const Station &y) { return x.x < y.x; }
inline bool compY(const Station &x, const Station &y) { return x.y < y.y; }

sort(stations.begin(), stations.end(), compX);
for (unsigned int i = 0; i < stations.size() - 1; ++i) {
for (unsigned int j = i + 1;
(stations[j].x == stations[i].x) and (j < stations.size()); ++j) {
addEdge(stations[i].id, stations[j].id,
abs(stations[i].y - stations[j].y));
}
}

sort(stations.begin(), stations.end(), compY);
for (unsigned int i = 0; i < stations.size() - 1; ++i) {
for (unsigned int j = i + 1;
(stations[j].y == stations[i].y) and (j < stations.size()); ++j) {
addEdge(stations[i].id, stations[j].id,
sabs(stations[i].x - stations[j].x));
}
}

跑 Dijkstra

  • 为了加速,我们使用堆优化的 Dijkstra
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inline bool operator<(const Edge &x, const Edge &y) { return x.len > y.len; }

array<unsigned int, MAX_M> dis;
bitset<MAX_M> vis;

fill(dis.begin(), dis.end(), INT32_MAX);

priority_queue<Edge> heap;
heap.push(Edge(0, 0, 0));
dis[0] = 0;

while (heap.size()) {
unsigned int now = heap.top().from;
heap.pop();

if (vis[now]) {
continue;
}
vis[now] = true;

for (auto i : graph[now]) {
if (dis[i.to] > dis[now] + i.len) {
dis[i.to] = dis[now] + i.len;
heap.push(Edge(i.to, 0, dis[i.to]));
}
}
}

最后,输出答案

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cout << dis[目标] << endl;

完整代码

  • 由于数据点没有 1-1 的情况,所以省略判断 1-1
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// Copyright (c) 2021 bmyjacks
// LastModified: 2021/7/1 下午7:24
// License: GPLv3
// Author: bmyjacks

#include <algorithm>
#include <array>
#include <bitset>
#include <cctype>
#include <cstdio>
#include <queue>
#include <vector>

using namespace std;

const unsigned int MAX_M = 1e5 + 2;

class Station {
public:
unsigned int id, x, y;
Station(const unsigned int &id, const unsigned int &x,
const unsigned int &y) {
this->id = id;
this->x = x;
this->y = y;
}
};
class Edge {
public:
unsigned int from, to;
unsigned int len;
Edge(const unsigned int &from, const unsigned int &to,
const unsigned int &len) {
this->from = from;
this->to = to;
this->len = len;
}
};

unsigned int n, m;
vector<Station> stations;
vector<Edge> graph[MAX_M];
array<unsigned int, MAX_M> dis;
bitset<MAX_M> vis;

inline void addEdge(const unsigned int &from, const unsigned int &to,
const unsigned int &len) {
graph[from].emplace_back(Edge(from, to, len));
graph[to].emplace_back(Edge(to, from, len));
}

inline bool operator<(const Edge &x, const Edge &y) { return x.len > y.len; }
inline bool compX(const Station &x, const Station &y) { return x.x < y.x; }
inline bool compY(const Station &x, const Station &y) { return x.y < y.y; }
inline unsigned int stationsDistance(const unsigned int &x,
const unsigned int &y) {
return x > y ? ((x - y) * 2 + 1) : ((y - x) * 2 + 1);
}

inline unsigned int read() {
unsigned int x = 0;
char ch = getchar();
while (!isdigit(ch)) {
ch = getchar();
}
while (isdigit(ch)) {
x = x * 10 + ch - 48;
ch = getchar();
}
return x;
}

void write(const unsigned int x) {
if (x > 9) {
write(x / 10);
}
putchar(x % 10 + 48);
}

int main() {
n = read();
m = read();

for (unsigned int i = 1, x, y; i <= m; ++i) {
x = read();
y = read();
stations.push_back(Station(i, x, y));
}

long long xFrom, yFrom, xTo, yTo;
xFrom = read();
yFrom = read();
xTo = read();
yTo = read();
stations.emplace_back(Station(0, xFrom, yFrom));
stations.emplace_back(Station(m + 1, xTo, yTo));

sort(stations.begin(), stations.end(), compX);
for (unsigned int i = 0; i < stations.size() - 1; ++i) {
for (unsigned int j = i + 1; stations[j].x == stations[i].x; ++j) {
addEdge(stations[i].id, stations[j].id,
stationsDistance(stations[i].y, stations[j].y));
}
}

sort(stations.begin(), stations.end(), compY);
for (unsigned int i = 0; i < stations.size() - 1; ++i) {
for (unsigned int j = i + 1; stations[j].y == stations[i].y; ++j) {
addEdge(stations[i].id, stations[j].id,
stationsDistance(stations[i].x, stations[j].x));
}
}

for (unsigned int i = 0; i <= m + 1; ++i) {
dis[i] = INT32_MAX;
}

priority_queue<Edge> heap;
heap.push(Edge(0, 0, 0));
dis[0] = 0;

while (heap.size()) {
unsigned int now = heap.top().from;
heap.pop();

if (vis[now]) {
continue;
}
vis[now] = true;

for (auto i : graph[now]) {
if (dis[i.to] > dis[now] + i.len) {
dis[i.to] = dis[now] + i.len;
heap.push(Edge(i.to, 0, dis[i.to]));
}
}
}

write(dis[m + 1] - 1);
putchar('\n');

return 0;
}