给一个n*n的图,0表示地面,1表示水,给出起点和终点,
现要从起点到达终点,有一次在两个坐标间创立隧道的机会,消耗为(x1 - x2)^2 + (y1 - y1)^2。
求出最小的消耗,如果直接能走到,则消耗为0。
bfs求出起点和终点分别能到达的点,枚举每一种情况,取最小值即可。
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
int Next[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
const int MAXN = 50 + 5;
char Map[MAXN][MAXN];
int vis[MAXN][MAXN];
int n, r1, r2, c1, c2;
int Bfs()
{
queue<pair<int, int>> que;
vector<pair<int, int>> start;
vector<pair<int, int>> ends;
que.push(make_pair(r1, c1));
vis[r1][c1] = 1;
while (!que.empty())
{
int x = que.front().first;
int y = que.front().second;
start.push_back(make_pair(x, y));
if (x == r2 && y == c2)
return 0;
for (int i = 0;i < 4;i++)
{
int nx = x + Next[i][0];
int ny = y + Next[i][1];
if (nx < 1 || nx > n || ny < 1 || ny > n)
continue;
if (vis[nx][ny] || Map[nx][ny] == ‘1‘)
continue;
vis[nx][ny] = 1;
que.push(make_pair(nx, ny));
}
que.pop();
}
que.push(make_pair(r2, c2));
vis[r2][c2] = 1;
while (!que.empty())
{
int x = que.front().first;
int y = que.front().second;
ends.push_back(make_pair(x, y));
for (int i = 0;i < 4;i++)
{
int nx = x + Next[i][0];
int ny = y + Next[i][1];
if (nx < 1 || nx > n || ny < 1 || ny > n)
continue;
if (vis[nx][ny] || Map[nx][ny] == ‘1‘)
continue;
vis[nx][ny] = 1;
que.push(make_pair(nx, ny));
}
que.pop();
}
int res = 9999;
/*
for (int i = 0;i < start.size();i++)
cout << start[i].first << ‘ ‘ << start[i].second << endl;
cout << endl;
for (int i = 0;i < ends.size();i++)
cout << ends[i].first << ‘ ‘ << ends[i].second << endl;
cout << endl;
*/
for (int i = 0;i < start.size();i++)
{
for (int j = 0;j < ends.size();j++)
{
int cx = start[i].first - ends[j].first;
int cy = start[i].second - ends[j].second;
int cost = cx * cx + cy * cy;
res = min(res, cost);
}
}
return res;
}
int main()
{
cin >> n;
cin >> r1 >> c1 >> r2 >> c2;
for (int i = 1;i <= n;i++)
{
for (int j = 1; j <= n; j++)
cin >> Map[i][j];
}
cout << Bfs() << endl;
return 0;
}
原文:https://www.cnblogs.com/YDDDD/p/10447796.html