技術標籤：最小生成樹

連線格點

題目

思路：把二維座標對映到一維的點上，就是一個最小生成樹問題，再把點連線，先連上下再連左右可以省去排序過程

具體程式碼如下

#include<iostream>
#include<algorithm>

using namespace std;

const int N = 1010, M = N*N, K = 2*M;

struct Node{
    int a, b, w;
}node[K];

int n, m, k;
int idx[N][N];
int p[M];

int find(int x){
    if(p[
 
x] != x) p[x] = find(p[x]);
    return p[x];
}

void get_edges()
{
    int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1}, dw[4] = {1, 2, 1, 2};

    for (int z = 0; z < 2; z ++ )
        for (int i = 1; i <= n; i ++ )
            for (int j = 1; j <= m; j ++ )
                for (int u = 0; u <
 
 4; u ++ )
                    if (u % 2 == z)
                    {
                        int x = i + dx[u], y = j + dy[u], w = dw[u];
                        if (x && x <= n && y && y <= m)
                        {
                            int a = idx[i][j], b =
 
 idx[x][y];
                            if (a < b) node[k ++ ] = {a, b, w};
                        }
                    }
}

int main(){
    
    cin >> n >> m;
    
    for(int i=1; i<=n*m; ++i) p[i] = i;
    for(int i=1, t=1; i<=n; ++i)
        for(int j=1; j<=m; ++j, ++t)
            idx[i][j] = t;
            
    int x1, x2, y1, y2;
    while(cin >> x1 >> y1 >> x2 >> y2){
        int a = idx[x1][y1], b = idx[x2][y2];
        p[find(a)] = find(b);
    }
    
    get_edges();
    
    int res = 0;
    for(int i=0; i<k; ++i){
        int a = node[i].a, b = node[i].b, w = node[i].w;
        a = find(a), b = find(b);
        if(a != b){
            p[a] = b;
            res += w;
        }
    }
    
    cout << res << endl;
    
    return 0;
}