UV 1151-buy or build poj 2784 buy or build (Minimum Spanning Tree)

It is also a simple Minimum Spanning Tree Algorithm.

However, some new things need to be added, which requires a deep understanding of the Minimum Spanning Tree Algorithm and the use of the query set.

The solution to the problem is also complicated.

#include<cstdio>#include<cstring>#include<vector>#include<algorithm>using namespace std;const int maxn = 1005;struct point{    int x;    int y;}pp[maxn];struct edge{    int s;    int e;    int dist;}l[maxn*maxn];int n,q,m;int p[maxn];vector<int> g[10];int c[10];int distance_(point a,point b){    return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);}int cmp(edge a,edge b){    return a.dist < b.dist;}int find_(int x){    return p[x]==x?x:p[x]=find_(p[x]);}bool merge_(int a,int b){    int x=find_(a);    int y=find_(b);    if(x==y) return false;    p[x]=y;    return true;}int kruskal(){    int ans=0;    int num=0;    for(int i=0;i<m&&num<n-1;i++)    {        if(merge_(l[i].s,l[i].e))        {            num++;            ans+=l[i].dist;        }    }    return ans;}void solve(){    for(int i=0;i<=n;i++) p[i]=i;    int ans = kruskal();    for(int s=1;s<(1<<q);s++)    {        int cost=0;        for(int tt=0;tt<=n;tt++) p[tt]=tt;        for(int j=0;j<q;j++)        {            if(!((s>>j)&1)) continue;            cost+=c[j];            for(int k=0;k<g[j].size();k++)            {                merge_(g[j][k],g[j][0]);            }        }        ans=min(ans,cost+kruskal());    }    printf("%d\n",ans);}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&q);        for(int i=0;i<10;i++) g[i].clear();        for(int i=0;i<q;i++)        {            int cnt;            scanf("%d%d",&cnt,&c[i]);            int a;            for(int j=0;j<cnt;j++)            {                scanf("%d",&a);                g[i].push_back(a);            }        }        for(int i=1;i<=n;i++)        {            scanf("%d%d",&pp[i].x,&pp[i].y);        }        m=0;        for(int i=1;i<=n;i++)        {            for(int j=i+1;j<=n;j++)            {                l[m].s=i;                l[m].e=j;                l[m++].dist=distance_(pp[i],pp[j]);            }        }        sort(l,l+m,cmp);        solve();        if(t) printf("\n");    }    return 0;}

The subset enumeration algorithm is required for the given solutions.

In the preceding solution, the binary help subset enumeration method is used, which is only applicable to the subset enumeration algorithm with small set elements.

The above method uses a struct to represent the edge, without opening so many arrays. I think struct can make the code more readable.

UV 1151-buy or build poj 2784 buy or build (Minimum Spanning Tree)