feat: 实现基于0-1 BFS的最大欧式距离计算

添加0-1 BFS算法来计算网格中可达点的最小代价,并在此基础上计算满足条件的两点间最大欧式距离。主要功能包括:
- 实现0-1 BFS算法计算各点到起点的最小代价
- 遍历所有点对,筛选满足代价条件的点对
- 计算并输出最大欧式距离
This commit is contained in:
Zengtudor 2025-10-01 21:02:08 +08:00
parent cbde29b00b
commit 75c2df61c9

View File

@ -1,3 +1,89 @@
int main(){
#include <cstdio>
#include <iostream>
#include <vector>
#include <deque>
#include <cmath>
#include <iomanip>
#include <algorithm>
using namespace std;
const int MAXN = 35;
const int dx[4] = {0, 1, 0, -1};
const int dy[4] = {1, 0, -1, 0};
int n, m, T;
char grid[MAXN][MAXN];
bool isValid(int x, int y) {
return x >= 0 && x < n && y >= 0 && y < m;
}
vector<vector<int>> bfs01(int sx, int sy) {
vector<vector<int>> dist(n, vector<int>(m, 0x3f3f3f3f));
vector<vector<bool>> visited(n, vector<bool>(m, false));
deque<pair<int, int>> dq;
}
int startcost = (grid[sx][sy] == '1') ? 1 : 0;
dist[sx][sy] = startcost;
dq.push_back({sx, sy});
while (!dq.empty()) {
auto [x, y] = dq.front();
dq.pop_front();
if (visited[x][y]) continue;
visited[x][y] = true;
for (int i = 0; i < 4; i++) {
int nx = x + dx[i];
int ny = y + dy[i];
if (!isValid(nx, ny)) continue;
int cost = (grid[nx][ny] == '1') ? 1 : 0;
if (dist[x][y] + cost < dist[nx][ny]) {
dist[nx][ny] = dist[x][y] + cost;
if (cost == 0) {
dq.push_front({nx, ny});
} else {
dq.push_back({nx, ny});
}
}
}
}
return dist;
}
double euladis(int x1, int y1, int x2, int y2) {
return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cin >> n >> m >> T;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
cin >> grid[i][j];
}
}
double maxdis = 0.0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
auto dist = bfs01(i, j);
for (int x = 0; x < n; x++) {
for (int y = 0; y < m; y++) {
if (dist[x][y] <= T) {
maxdis = max(maxdis, euladis(i, j, x, y));
}
}
}
}
}
printf("%.6f\n", maxdis);
}