POJ - 2349 Arctic Network (prim裸题)

POJ - 1287 Networking

题目

求最小生成树第s大的边

分析

最小生成树直接prim，用链式前向星存图。
不过prim比较适合邻接矩阵

#include 
#include 
#include 
#include 
#include 
#define INF 0x3f3f3f3f
#define d(x) cout << (x) << endl
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
using namespace std;
typedef long long ll;
const int mod = 1e9 + 7;
const int N = 5e2 + 10;

struct point{
    double x, y;
} a[N];

struct node{
    int to, next;
    double len;
} e[N * N];
int head[N * N];
int cnt;

int t, s, p, num;
double ans[N];

void add(int u, int v, double l)
{
    e[cnt].to = v;
    e[cnt].next = head[u];
    e[cnt].len = l;
    head[u] = cnt++;
}

void init()
{
    num = cnt = 0;
    memset(head, -1, sizeof(head));
}

double dis(int u, int v)
{
    return sqrt((a[u].x - a[v].x) * (a[u].x - a[v].x) + (a[u].y - a[v].y) * (a[u].y - a[v].y));
}

void prim()
{
    double dist[N];
    int k;
    memset(dist, INF, sizeof(dist));
    for (int i = head[1]; i != -1; i = e[i].next){
        dist[e[i].to] = e[i].len;
    }
    for (int i = 1; i <= p; i++){
        double min = INF;
        for (int j = 1; j <= p; j++){       //找最小距离
            if (dist[j] != 0 && dist[j] < min){
                min = dist[j];
                k = j;
            }
        }
        ans[num++] = min;
        dist[k] = 0;            //访问过就标记
        k = head[k];
        while(k != -1){         //用第k节点更新dist
            if(dist[e[k].to] != 0 && e[k].len < dist[e[k].to]){
                dist[e[k].to] = e[k].len;
            }
            k = e[k].next;
        }
    }
}

int main()
{
    for (scanf("%d", &t); t; t--){
        init();
        scanf("%d%d", &s, &p);
        for (int i = 1; i <= p; i++){
            scanf("%lf%lf", &a[i].x, &a[i].y);
            for(int j = 1; j < i; j++){
                add(i, j, dis(i, j));
                add(j, i, dis(i, j));
            }
        }
        prim();
        sort(ans, ans + num);
        printf("%.2f\n", ans[p - s]);
    }
    return 0;
}