hdu1102&hdu4081最小生成树，次小生成树

hdu1102
MST模板题
复习下kruskal，补一下之前偷懒没学的prim

赤裸裸的kruskal（lrj风格

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int N=107;
int n,m,g[N][N];
struct Edge{
    int x,y,w;
    Edge():x(0),y(0),w(0){}
    Edge(int _x,int _y,int _w):x(_x),y(_y),w(_w){}
    bool operator < (const Edge &a)const{return w<a.w;}
}edges[N*N];
int home[N];//²¢²é¼¯
int Find(int x){return home[x]==x?x:home[x]=Find(home[x]);}

int kruskal(){
    sort(edges+1,edges+m+1);
    for (int i=1;i<=n;i++)home[i]=i;
    int sum=0,cnt=0;
    for (int i=1;i<=m;i++){
        int x=Find(edges[i].x);
        int y=Find(edges[i].y);
        if (x==y)continue;
        home[x]=y;
        sum+=edges[i].w;
        cnt++;
        if (cnt==n-1)break;
    }
    return sum;
}

int main(){
    for (;scanf("%d",&n)==1;){
        for (int i=1;i<=n;i++){
            for (int j=1;j<=n;j++){
                scanf("%d",&g[i][j]);
            }
        }
        int x,y,q;
        scanf("%d",&q);
        for (;q--;){
            scanf("%d%d",&x,&y);
            g[x][y]=g[y][x]=0;
        }
        m=0;
        for (int i=1;i<=n;i++){
            for (int j=1;j<i;j++){
                edges[++m]=Edge(i,j,g[i][j]);
            }
        }
        printf("%d
",kruskal());
    }
    return 0;
}

赤裸裸的prim
看了巫泽俊的板（《挑战》书里

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int N=107;
int n,g[N][N];

int prim(){
    int minw[N];//MinWeight
    bool used[N];
    memset(used,0,sizeof(used));
    memset(minw,0x7f,sizeof(minw));
    minw[1]=0;
    int sum=0;
    while (1){
        int v=-1;
        for (int i=1;i<=n;i++){
            if (!used[i]&&(v==-1||minw[i]<minw[v]))v=i;
        }
        if (v==-1)break;
        used[v]=1;
        sum+=minw[v];
        for (int i=0;i<=n;i++){
            minw[i]=min(minw[i],g[v][i]);
        }
    }
    return sum;
}

int main(){
    for (;scanf("%d",&n)==1;){
        for (int i=1;i<=n;i++){
            for (int j=1;j<=n;j++){
                scanf("%d",&g[i][j]);
            }
        }
        int x,y,q;
        scanf("%d",&q);
        for (;q--;){
            scanf("%d%d",&x,&y);
            g[x][y]=g[y][x]=0;
        }
        printf("%d
",prim());
    }
    return 0;
}

为何我感觉还是kruskal简单呢，因为无脑贪心吗。。
补prim是为了这题
hdu4081
秦始皇啊，，，
这里有篇题解很详细的讲了次小生成树算法的证明和代码，
看完中文讲解，自己强行在上面的prim代码里面改，改的亲爹都不认识了，调了半天，过了样例，WA了，，，又对照题解里代码改，改的似乎亲爹认识了，，，依然WA，，，心累，玩个游戏叫找不同 = =终于A了，，，代码还是写的少了
废话了一大堆，上代码

#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
const int N=1007;
int n;
double g[N][N],maxw[N][N];
bool used[N][N];
struct City{
    int x,y,w;//w=population
    City():x(0),y(0),w(0){}
    City(int _x,int _y,int _w):x(_x),y(_y),w(_w){}
}citys[N];
double dist(City a,City b){
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

double prim(){
    int from[N];
    bool vis[N];
    double minw[N];
    for(int i=1;i<=n;i++){
        minw[i]=g[1][i];
        from[i]=1;
    }
    memset(used,0,sizeof(used));
    memset(maxw,0,sizeof(maxw));
    memset(vis,0,sizeof(vis));
    vis[1]=1;
    minw[1]=0;
    double sum=0;
    while (1){
        int v=-1;
        for (int i=1;i<=n;i++)if (!vis[i]){
            if (v==-1||minw[i]<minw[v])v=i;
        }
        if (v==-1)break;
        //printf("from[%d]=%d
",v,from[v]);
        used[v][from[v]]=used[from[v]][v]=1;
        vis[v]=1;
        sum+=minw[v];
        //printf("minw[%d]=%.8lf
",v,minw[v]);
        for (int i=1;i<=n;i++){
            if (!vis[i]&&g[v][i]<minw[i]){
                minw[i]=g[v][i];
                from[i]=v;
            }
            if (vis[i]&&i!=v){
                maxw[i][v]=maxw[v][i]=max(maxw[i][from[v]],minw[v]);
                //printf("    maxw[%d][%d]=%.8lf
",i,v,maxw[i][v]);
            }
        }
    }
    //for (int i=1;i<=n;i++)printf("from[%d]=%d
",i,from[i]);
    return sum;
}

int main(){
    //freopen("fuck.in","r",stdin);
    int T;scanf("%d",&T);
    for (;T--;){
        scanf("%d",&n);int x,y,z;
        for (int i=1;i<=n;i++){
            scanf("%d%d%d",&x,&y,&z);
            citys[i]=City(x,y,z);
        }
        for (int i=1;i<=n;i++)for (int j=1;j<=n;j++)
            g[i][j]=dist(citys[i],citys[j]);

        double sum=prim(),ans=-1;
        for (int i=1;i<=n;i++){
            for (int j=i+1;j<=n;j++){
                if (used[i][j])ans=max(ans,(citys[i].w+citys[j].w)/(sum-g[i][j]));
                else ans=max(ans,(citys[i].w+citys[j].w)/(sum-maxw[i][j]));
            }
        }
        printf("%.2lf
",ans);
    }
    return 0;
}

大家晚安，
我的心愿是
！！世界和平！！