UVA10397 - Connect the Campus(最小生成树)
题目大意:给你n个点,然后再给你m个已经连接的边,问如何使得所有的点都相连并且新建的边长度之和最小。
解题思路:最小生成树,但是有m条边是已经建好的,就将这些边的权值变成0,然后用kruskal的方法来求长度。
代码:
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 755;
const int maxm = 3e5;
struct Node {
    int x, y;
}node[maxn];
double w[maxm];
int p[maxn], u[maxm], v[maxm];
int vis[maxn][maxn], r[maxm];
double dist (int i, int j) {
    double x = node[i].x - node[j].x;
    double y = node[i].y - node[j].y;
    return sqrt(x * x + y * y);
}
void init (int n) { 
    for (int i = 0; i < n; i++)
        p[i] = i;
}
int getParent (int x) {
    return (x == p[x]) ? p[x] : p[x] = getParent(p[x]);
}
int cmp (int i, int j) {
    return w[i] < w[j];
}
double Kruskal (int n, int m) {
    double ans = 0;
    init(n);
    for (int i = 0; i < m; i++)
        r[i] = i;
    sort(r, r + m, cmp);
    for (int i = 0; i < m; i++) {
        int P = getParent(u[r[i]]);
        int Q = getParent(v[r[i]]);
        if (P == Q) continue;
        ans += w[r[i]];
        p[P] = Q;
    }
    return ans;
}
int main () {
    int n, m;
    while (scanf ("%d", &n) != EOF) {
        for (int i = 0; i < n; i++) 
            scanf ("%d%d", &node[i].x, &node[i].y); 
        int x, y;
        memset (vis, 0, sizeof(vis));
        scanf ("%d", &m);   
        for (int i = 0; i < m; i++) {
            scanf ("%d%d", &x, &y);
            vis[x - 1][y - 1] = vis[y - 1][x - 1] = 1;
        }
        int pos = 0;
        for (int i = 0; i < n; i++)
            for (int j = i + 1; j < n; j++) {
                u[pos] = i;
                v[pos] = j;
                if (vis[i][j])
                    w[pos++] = 0;
                else
                    w[pos++] = dist(i, j);
            }
        printf ("%.2lf\n", Kruskal(n, pos));    
    }
    return 0;
}UVA10397 - Connect the Campus(最小生成树+并查集)
原文:http://blog.csdn.net/u012997373/article/details/44132403