hdu 4081 次小生成树

链接

http://acm.hdu.edu.cn/showproblem.php?pid=4081

题意

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找到一个生成树，并且保证A/B最大，所以B要尽量小， A尽量大。

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解析

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B尽量小，就要想到最小生成树，但是最小生成树里面，可以删除一个边，这样就转化为次小生成树中找A/B最大的情况。或者prim+并查集。

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代码1

运行时间：46ms

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
#define INF 0x3f3f3f3f
struct node
{
    int a,b;
    int num;
};
struct nn
{
    int a,b;
    double val;
    bool operator <(const nn &d) const {
        return val <d.val;
    }
};
node A[1010];
nn B[1010];
int n;
double dis[1010];
bool vis[1010];
int num[1010];
int pre[1010];
int fa[1010];
int M[1010];
int mm;
void init()
{
    for(int i=1;i<=n;i++)
        fa[i]=i,M[i]=A[i].num;
}
int fi(int a){ return a==fa[a] ?  a:(fa[a]=fi(fa[a]));}
void un(int a,int b) { a=fi(a),b=fi(b); fa[a]=b; M[b]=max(M[b],M[a]);}
void work()
{
    int i,j;
    memset(vis,false,sizeof vis);
    memset(pre,-1,sizeof pre);
    for(i=0;i<=n;i++)
        dis[i]=INF;
    dis[1]=0;
    double ans=0;
    double sum=0;
    int m=0;
    while(1)
    {
        int v=-1;
        for(i=1;i<=n;i++)
        {
            if(!vis[i]&&(v==-1||dis[i]<dis[v])) v=i;
        }
        if(v==-1) break;
        vis[v]=true;
        sum=sum+dis[v];
        if(pre[v]!=-1) B[++m].a=pre[v],B[m].b=v,B[m].val=dis[v];
        for(i=1;i<=n;i++)
        {
            double t=sqrt((double)((A[v].a-A[i].a)*(A[v].a-A[i].a)+(A[v].b-A[i].b)*(A[v].b-A[i].b)));
            if(t<dis[i]) dis[i]=t,pre[i]=v;
        }
    }
    sort(B+1,B+1+m);
    init();
    for(i=1;i<=n;i++)
    {
        int a=fi(B[i].a);
        int b=fi(B[i].b);
    //    cout<<B[i].a<<" "<<B[i].b<<" "<<M[a]<<" "<<M[b]<<endl;
        ans=max(ans,(double)(M[a]+M[b])/(sum-B[i].val));
        un(a,b);
    }
    printf("%.2lf\n",ans);
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int i,j;
        scanf("%d",&n);
        for(i=1;i<=n;i++)
            scanf("%d%d%d",&A[i].a,&A[i].b,&A[i].num);
        work();
    }
    return 0;
}

代码2：时间复杂度：156ms

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
#include <map>
#include <cmath>
using namespace std;
const int inf=0x3f3f3f3f;
const int maxn = 1000+100;
typedef long long LL;
int fa[maxn], head[maxn];
int en[maxn];
double length[maxn][maxn];
struct Edge{
    int t;
    int next;
    Edge (int _t, int _n) {
        t = _t;
        next = _n;
    }
    Edge(){}
}e[500000+10];
int cnt;

void add(int u, int v) {
    e[cnt] = Edge(v, head[u]);
    head[u] = cnt;
    en[u] = cnt;
    cnt++;
}

void init() {
    for (int i=0; i<maxn; i++)
        fa[i] = i;
    cnt = 0;
    memset(head,-1, sizeof(head));
}

struct point {
    double x, y;
    int pu;
    bool operator < (const point &a) const  {
        if (x == a.x)
            return y<a.y;
        return x<a.x;
    }
}p[maxn];

int Find(int x) {
    return x==fa[x]?x:fa[x]=Find(fa[x]);
}

void Union(int u, int v) {
    u = Find(u);
    v = Find(v);
    if (u != v)
        fa[u] = v;
}

struct node {
    int u, v;
    double d;
    double pu;
    bool operator < (const node &a) const {
        return d<a.d;
    }
}edge[1000000+10];

double Dis(point a, point b) {
    return sqrt((a.x-b.x)*(a.x-b.x) + (b.y-a.y)*(b.y-a.y));
}

int main()
{
    int T;
    scanf("%d", &T);
    while (T--) {
        init();
        int n;
        scanf("%d", &n);
        for (int i=0; i<n; i++)
            scanf("%lf%lf%d", &p[i].x, &p[i].y, &p[i].pu);
        for (int i=0; i<n; i++) 
            add(i, i);
        int m=0;
        sort(p, p+n);
        for (int i=0; i<n-1; i++) {
            for (int j=i+1; j<n; j++) {
                double dis = Dis(p[i], p[j]);
                edge[m].u = i;
                edge[m].v = j;
                edge[m].d = dis;
                edge[m].pu = double (p[i].pu+p[j].pu);
                m++;
            }
        }
        sort(edge, edge+m);
        double DIS = 0;
        for (int i=0; i<m; i++) {
            int u = Find(edge[i].u);
            int v = Find(edge[i].v);
            if (u == v)
                continue;
            for (int j=head[u]; j!=-1; j=e[j].next) {
                for (int k=head[v]; k!=-1; k=e[k].next) {
                    length[e[j].t][e[k].t] = length[e[k].t][e[j].t] = edge[i].d;
                }
            }
            e[en[v]].next = head[u];
            en[v] = en[u];
            Union(u, v);
            DIS += edge[i].d;
        }

        double maxx = 0;
        for (int i=0; i<m; i++) {
            double b = edge[i].pu/(DIS-length[edge[i].u][edge[i].v]);
            if (b > maxx)
                maxx = b;
        }
        printf("%.2lf\n", maxx);
    }
    return 0;
}