HDU - 1875 （最小生成树）

HDU - 1875 （最小生成树）

题目链接

思路：把每个顶点编号，点之间的距离>=10且<=100,给这个两个点之间建边。建完之后求求最小生成树（MST）

AC代码

#include 
#include 
#include 
using namespace std;
const int N = 110;
struct node{
    int u, v;
    double w;
}q[N*N];
int n, m, cnt;
int xx[N], yy[N], pre[N];
void init(){
    for(int i = 1; i <= n; i++){
        pre[i] = i;
    }
    cnt = 0;
}
int Find(int x){
    if(x != pre[x]) pre[x] = Find(pre[x]);
    return pre[x];
}
void join(int x, int y){
    int tx = Find(x);
    int ty = Find(y);
    if(tx != ty) pre[ty] = tx;
}
bool cmp(node a, node b){
    return a.w < b.w;
}
double Kru(){
    double ans = 0;
    sort(q, q + cnt, cmp);
    for(int  i = 0; i < cnt; i++){
        if(Find(q[i].u) != Find(q[i].v)){
            join(q[i].u, q[i].v);
            ans += q[i].w;
        }
    }
    return ans;
}
int main(){
    #ifdef ONLINE_JUDGE
    #else
        freopen("in.txt", "r", stdin);
    #endif //ONLINE_JUDGE
    int t;
    scanf("%d", &t);
    while(t--){
        scanf("%d", &n);
        init();
        for(int  i = 1; i <= n; i++){
            scanf("%d%d", &xx[i], &yy[i]);
        }
        for(int i = 1; i <= n; i++){
            for(int j = 1; j <= n; j++){

                int dis = (xx[i] - xx[j])*(xx[i] - xx[j])+(yy[i] - yy[j])*(yy[i] - yy[j]);
                if(dis >= 100.0 && dis <= 1000000.0) {
                    q[cnt].u = i;
                    q[cnt].v = j;
                    q[cnt++].w = sqrt((double)dis);
                }
            }
        }
        double ans = Kru();
        int cnt1 = 0;
        for(int i = 2; i <= n; i++){
            if(pre[i] == i) cnt1++;
        }
        if(cnt1 > 1) printf("oh!\n");
        else  printf("%.1lf\n", ans*100.0);
    }
    return 0;
}

posted on 2018-10-23 18:48 坤sir 阅读(...) 评论(...) 编辑 收藏