HDU 1102 Constructing Roads

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题目大意：给你n*n的一个临界矩形每个数代表的是当前i行与j列的距离之后给你Q个连通的道路以下Q行为哪两个村庄连通求最小连通的距离也就是求最小生成树

解题思路：这道题甚至已经把临界矩形都告诉你了比模板题更要水两种for循环输入连通的道路距离标记为0

普利姆算法

//Memory: 252 KB   //Time: 16 MS  #include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#include <map>#include <cmath>#include <queue>#include <string>#include <vector>#include <set>using namespace std;#define ll long long#define sc(x) scanf("%d",&x)#define dsc(x,y) scanf("%d%d",&x,&y)#define pr(x) printf("%d\n",x)#define FOR(i,n,o) for(int i=o;i<=n;i++)#define lcr(a,b)  memset(a,b,sizeof(a))#define Inf 1<<29#define N 104int n,q,a,b;int mp[N][N],vis[N],dis[N],cis[N][N];int pri(){         int sum=0;         FOR(i,n,1)         {             dis[i]=mp[1][i];         }         vis[1]=1;         FOR(i,n-1,1)         {             int to=-1;             int minn=Inf;             FOR(j,n,1)             {             if(!vis[j]&&dis[j]<minn)              {                  to=j;                  minn=dis[j];              }         }         if(to==-1)          return 0;          vis[to]=1;          sum+=minn;          FOR(j,n,1)          {              if(!vis[j])                dis[j]=min(dis[j],mp[to][j]);          }         }         return sum;}int main(){    while(~sc(n))    {        lcr(vis,0);        lcr(cis,0);        FOR(i,n,1)        {            FOR(j,n,1)            {                mp[i][j]=Inf;            }        }        FOR(i,n,1)        {           FOR(j,n,1)           {               int v;            scanf("%d",&v);            mp[i][j]=v;           }        }        sc(q);        while(q--)        {            dsc(a,b);            mp[a][b]=0;            mp[b][a]=0;        }        pr(pri());    }    return 0;}

克鲁斯卡尔算法

//Memory: 332 //Time: 79 #include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#include <map>#include <cmath>#include <queue>#include <string>#include <vector>#include <set>using namespace std;#define ll long long#define sc(x) scanf("%d",&x)#define dsc(x,y) scanf("%d%d",&x,&y)#define pr(x) printf("%d\n",x)#define FOR(i,n,o) for(int i=o;i<=n;i++)#define lcr(a,b)  memset(a,b,sizeof(a))#define Inf 1<<29#define N 105int n,v,p[N],mp[N][N];struct node{    int x;    int y;    int l;}q[N*N];int Find(int x){     if(x!=p[x])     {         p[x]=Find(p[x]);     }     return p[x];}int cmp(node a,node b){    return a.l<b.l;}int Kruskal(int l){      int sum=0;      int to=0;      FOR(i,l,1)      {          int xx=Find(q[i].x);          int yy=Find(q[i].y);          if(xx!=yy)          {              p[yy]=xx;              sum+=q[i].l;              to++;              if(to==n-1)                return sum;          }      }      return sum;}int main(){    while(~sc(n))    {        int k=1;        FOR(i,n,1)        {            p[i]=i;            FOR(j,n,1)            {                sc(v);               mp[i][j]=v;            }        }        int m;        sc(m);        while(m--)        {            int a,b;            scanf("%d %d",&a,&b);            mp[a][b]=0;            mp[b][a]=0;        }        FOR(i,n,1)        {            FOR(j,n,1)            {                q[k].x=i;                q[k].y=j;                q[k++].l=mp[i][j];            }        }        sort(q+1,q+k,cmp);        int ans=Kruskal(k);        pr(ans);    }    return 0;}

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