kuangbin专题六 最小生成树从入门到熟练5题

Posted 2021-02-02 zhenghanghu

POJ 2031 Building a Space Station

猜一个结论两个球体间的最短距离是圆心间距离减去两个球的半径，如果是负数就说明相交。

然后理解了三维上两个点之间的距离怎么求

//#include<bits/stdc++.h>
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<map>
#include<cmath>
#include<vector>
#define inf 2e9
#define maxnode 200000
#define ll long long
#define lowbit(x) (x&(-x))
const int mod = 1e9 + 7;
const int maxn = 1e5 + 10;
int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0};
using namespace std;

int par[105],cnt;
struct edge{
    int u,v;
    double cost;
    edge(int u1=0,int v1=0,double c1=0): u(u1),v(v1),cost(c1) {}
}edges[10010];
bool cmp(edge n1,edge n2){
    return n1.cost<n2.cost;
}

int find_root(int x){
    if( x==par[x] ) return x;
    return par[x] = find_root( par[x] );
}

double x[105],y[105],z[105],r[105];

double find_dist(int i,int j){
    //圆心间距离
    double dis = sqrt( abs(x[i]-x[j])*abs(x[i]-x[j]) + abs(y[i]-y[j])*abs(y[i]-y[j]) + abs(z[i]-z[j])*abs(z[i]-z[j]) );
    dis-=r[i]; dis-=r[j];
    return dis;
}

int main(){
    int n;
    while(1){
        cin>>n; if(n==0) break;

        cnt=0;
        for(int i=1;i<=n;i++) par[i]=i;

        for(int i=1;i<=n;i++) cin>>x[i]>>y[i]>>z[i]>>r[i];
        for(int i=1;i<=n;i++){
            for(int j=i+1;j<=n;j++){//find distance
                double dist = find_dist(i,j);
                if( dist<=0 ){//i跟j一开始就在一个集合
                    int rooti = find_root(i), rootj = find_root(j);
                    par[rooti] = rootj;
                }
                else edges[++cnt] = edge(i,j,dist);
            }
        }
        sort(edges+1,edges+1+cnt,cmp);
        //for(int i=1;i<=cnt;i++) cout<<edges[i].u<<" "<<edges[i].v<<" "<<edges[i].cost<<endl;
        double ans=0;
        for(int i=1;i<=cnt;i++){
            int u = edges[i].u,v=edges[i].v;
            int rootu = find_root(u), rootv = find_root(v);
            if( rootu!=rootv  ){
                par[rootu] = rootv;
                ans+=edges[i].cost;
            }
        }
        printf("%.3lf
",ans);

    }


    return 0;
}

View Code

 

POJ 1789 Truck History

如果把每个Truck当作一个顶点的话，那任意一颗生成树都能对应上一个derive的方案，那就是求MST

//#include<bits/stdc++.h>
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<map>
#include<cmath>
#include<vector>
#define inf 2e9
#define maxnode 200000
#define ll long long
#define lowbit(x) (x&(-x))
const int mod = 1e9 + 7;
const int maxn = 2e3 + 10;
int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0};
using namespace std;

int dist[maxn],vis[maxn];//从mst集合到这个点的距离

int a[maxn][maxn];
char trucks[maxn][10];
int find_difference(int i,int j){
    int ans=0;
    for(int k=0;k<7;k++){
        if( trucks[i][k]!=trucks[j][k] ) ans++;
    }
    return ans;
}

int main(){
    int n;
    while( scanf("%d",&n)){
        if(n==0) break;

        for(int i=1;i<=n;i++) scanf("%s",trucks[i]);
        for(int i=1;i<=n;i++){
            for(int j=i+1;j<=n;j++){
                a[i][j] = a[j][i] = find_difference(i,j);
            }
        }
        memset(vis,0,sizeof(vis));
        //mst
        int ans=0;
        for(int i=2;i<=n;i++){
            dist[i] = a[1][i];
        }

        for(int i=2;i<=n;i++){
            int mind=10,index;
            for(int j=2;j<=n;j++){
                if( vis[j]==0 && dist[j]<mind ) { mind=dist[j]; index=j; }
            }
            //
            ans+=dist[index]; vis[index]=1;
            for(int j=2;j<=n;j++){
                if( j!=index && dist[j]>a[index][j] ) dist[j] = a[index][j];
            }
        }

        printf("The highest possible quality is 1/%d.
",ans);

    }
    


    return 0;
}

View Code

 

POJ 2349 Arctic Network

知道结论：所有最小生成树边按边的权值排序后，序列一样

找第k大的边就行了

//#include<bits/stdc++.h>
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<map>
#include<cmath>
#include<vector>
#define inf 2e9
#define maxnode 200000
#define ll long long
#define lowbit(x) (x&(-x))
const int mod = 1e9 + 7;
const int maxn = 5e2 + 10;
int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0};
using namespace std;

int a[maxn][maxn],dist[maxn];
int x[maxn],y[maxn],vis[maxn];

int find_dist(int i,int j){
    return abs(x[i]-x[j])*abs(x[i]-x[j])+abs(y[i]-y[j])*abs(y[i]-y[j]);
}

int main(){
    int t; scanf("%d",&t);
    while( t-- ){
        int s,n; scanf("%d%d",&s,&n);
        for(int i=1;i<=n;i++) scanf("%d%d",x+i,y+i);
        for(int i=1;i<=n;i++){
            for(int j=i+1;j<=n;j++){
                a[i][j] = a[j][i] = find_dist(i,j);
             //   cout<<i<<" "<<j<<" "<<a[i][j]<<endl;
            }
        }
        memset(vis,0,sizeof(vis));
        //mst
        for(int i=2;i<=n;i++) dist[i] = a[1][i];
        vector<int> edge; edge.clear();

        for(int i=2;i<=n;i++){
            int mind=inf,index;
            for(int j=2;j<=n;j++){ if( vis[j]==0 && dist[j]<mind ) { mind=dist[j]; index=j; } }
           // cout<<"!!! "<<index<<" "<<dist[index]<<endl;
            edge.push_back(dist[index]); vis[index]=1;
            for(int j=2;j<=n;j++){ if( j!=index && dist[j]>a[index][j] ) dist[j] = a[index][j]; }
        }

       // for(int i=0;i<edge.size();i++) cout<<edge[i]<<" "; cout<<endl;
        sort(edge.begin(),edge.end()); reverse(edge.begin(),edge.end());
       // for(int i=0;i<edge.size();i++) cout<<edge[i]<<" "; cout<<endl;
        printf("%.2lf
",sqrt( double(edge[ s-1  ]) ) );
    }
    


    return 0;
}

View Code

 

POJ 3026 Borg Maze

bfs + kruskal

主要是建模

//#include<bits/stdc++.h>
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<queue>
#include<cstring>
#include<map>
#include<vector>
#define inf 2e9
#define maxnode 200000
#define ll long long
#define lowbit(x) (x&(-x))
const int mod = 1e9 + 7;
const int maxn = 5e2 + 10;
int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0};
using namespace std;

int par[maxn],cnt,mp[maxn][maxn];
struct edge{
    int u,v,cost;
    edge(int u1=0,int v1=0,int c1=0): u(u1),v(v1),cost(c1) {}
}edges[maxn*maxn];
bool cmp(edge n1,edge n2){
    return n1.cost<n2.cost;
}

int find_root(int x){
    if( x==par[x] ) return x;
    return par[x] = find_root( par[x] );
}

int x[maxn],y[maxn],vis[maxn][maxn],n,m,mat[maxn][maxn];
char a[maxn][maxn];

struct node{
    int x,y,step;
    node(int x1=0,int y1=0,int s1=0): x(x1),y(y1),step(s1) {}
};

void bfs(int s){
    queue<node> q; 
    memset(vis,0,sizeof(vis));
    q.push( node(x[s],y[s],0) ); vis[ x[s] ][ y[s] ]=1;
    while( !q.empty() ){
        node u=q.front(); q.pop();
        if( mp[u.x][u.y] ) mat[s][ mp[u.x][u.y] ] = u.step;
        for(int k=0;k<4;k++){
            int x1=u.x+dx[k];
            int y1=u.y+dy[k];
            if( x1>=1 && x1<=n && y1>=1 && y1<=m && a[x1][y1]!=‘#‘ && vis[x1][y1]==0 ){
                vis[x1][y1]=1;
                q.push( node(x1,y1,u.step+1) );
            }
        }
    }

}

int main(){
    int t,tc=0; scanf("%d",&t);
    while( t-- ){

        scanf("%d%d",&m,&n); char temp[51]; gets(temp);
        memset(mp,0,sizeof(mp));
        cnt=0;

        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                scanf("%c",&a[i][j]);
                if( a[i][j]==‘S‘ || a[i][j]==‘A‘ ) { x[++cnt]=i; y[cnt]=j; mp[i][j]=cnt; }
            }
            getchar();
        }

        for(int i=1;i<=cnt;i++) bfs(i);
        for(int i=1;i<=cnt;i++) par[i]=i;
        
        int cnt_edge=0;
        for(int i=1;i<=cnt;i++)
            for(int j=i+1;j<=cnt;j++) edges[++cnt_edge] = edge(i,j, mat[i][j] );

        sort(edges+1,edges+1+cnt_edge,cmp);

        int ans=0;
        for(int i=1;i<=cnt_edge;i++){
            int u = edges[i].u,v=edges[i].v;
            int rootu = find_root(u), rootv = find_root(v);
            if( rootu!=rootv  ){
                par[rootu] = rootv;
                ans+=edges[i].cost;
            }
        }

        printf("%d
",ans);
    }


    return 0;
}

View Code

 

POJ 1679 The Unique MST

次小生成树

求出数组mat[u][v]表示最小生成树上u到v路径上的最长边边权，用dp的思路进行不断迭代，考验一定码力

//#include<bits/stdc++.h>
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<map>
#include<cmath>
#include<vector>
#define inf 2e9
#define maxnode 200000
#define ll long long
#define lowbit(x) (x&(-x))
const int mod = 1e9 + 7;
const int maxn = 1e2 + 10;
int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0};
using namespace std;

int mp[maxn][maxn],dist[maxn],pre[maxn];
int mat[maxn][maxn],used[maxn][maxn],vis[maxn];//mat[u][v] u到v最短路径的最长距离

int main(){
    int t; cin>>t;
    while( t-- ){
        int n,m; cin>>n>>m;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++) mp[i][j]=inf;
        memset(used,0,sizeof(used));
        memset(mat,0,sizeof(mat));
        memset(vis,0,sizeof(vis));
        

        for(int i=1;i<=m;i++){
            int u,v,d; cin>>u>>v>>d;
            mp[u][v]=mp[v][u]=d;
        }

        vis[1]=1;
        for(int i=2;i<=n;i++) { dist[i]=mp[1][i]; pre[i]=1; }

        int ans=0;
        for(int i=2;i<=n;i++){
            int mind=inf,index;
            for(int j=2;j<=n;j++){
                if( !vis[j] && dist[j]<mind ) { mind=dist[j]; index=j; } 
            }
            //找到要加入vis的顶点index
            used[index][pre[index]] = used[pre[index]][index]= 1;
            vis[index]=1;
            ans+=dist[index];
            for(int j=2;j<=n;j++){
                if( vis[j] ) mat[j][index] = mat[index][j] = max( mat[j][pre[index]],dist[index] );
                if( !vis[j] && mp[index][j]<dist[j] ){
                    dist[j]=mp[index][j];
                    pre[j]=index;
                }
            }

        }
        //mst长度，并且mat维护出来
        int minlen=inf;
        for(int i=1;i<=n;i++){
            for(int j=i+1;j<=n;j++){
                if( used[i][j]==0 && mp[i][j]!=inf ){
                    minlen = min( minlen,ans+mp[i][j]-mat[i][j] );
                }
            }
        }
        if( minlen==ans ) cout<<"Not Unique!"<<endl;
        else cout<<ans<<endl;

    }
    


    return 0;
}

View Code