poj2421

  1 //Accepted    312 KB    63 ms    
  2 //kruskal算法，求最小生成树，把已经存在边的两点之间的权值置为0； 
  3 #include <cstdio>
  4 #include <cstring>
  5 #include <algorithm>
  6 using namespace std;
  7 const int MAXN = 105;
  8 const int inf = 100000000;
  9 const int MAXM =MAXN*MAXN;
 10 struct node
 11 {
 12     int u,v,c;
 13     node(int u,int v,int c):u(u),v(v),c(c)
 14     {
 15         
 16     }
 17     node()
 18     {
 19         
 20     }
 21 }p[MAXM];
 22 int cmp(node x,node y)
 23 {
 24     return x.c<y.c;
 25 }
 26 bool flag[MAXN][MAXN];
 27 int a[MAXN][MAXN];
 28 int n;
 29 int Q;
 30 int e;
 31 int f[MAXN];
 32 int ans;
 33 void addnode(int u,int v,int c)
 34 {
 35     p[e++]=node(u,v,c);
 36 }
 37 void build()
 38 {
 39     for (int i=0;i<=n;i++)
 40     {
 41         f[i]=i;
 42     }
 43 }
 44 int find(int x)
 45 {
 46     if (f[x]==x) return x;
 47     return f[x]=find(f[x]);
 48 }
 49 void union_set(int x,int y)
 50 {
 51     int fx=find(x),fy=find(y);
 52     if (fx!=fy)
 53     {
 54         f[fy]=fx;
 55     }
 56 }
 57 void kruskal()
 58 {
 59     int cnt=0;
 60     ans=0;
 61     sort(p,p+e,cmp);
 62     for (int i=0;i<e;i++)
 63     {
 64         int x=p[i].u;
 65         int y=p[i].v;
 66         if (find(x)!=find(y))
 67         {
 68             union_set(x,y);
 69             ans+=p[i].c;
 70             cnt++;
 71         }
 72         if (cnt==n-1) return ;
 73     }
 74 }
 75 int main()
 76 {
 77     while (scanf("%d",&n)!=EOF)
 78     {
 79         build();
 80         for (int i=0;i<n;i++)
 81         for (int j=0;j<n;j++)
 82         scanf("%d",&a[i][j]);
 83         scanf("%d",&Q);
 84         e=0;
 85         memset(flag,0,sizeof(flag));
 86         for (int i=0;i<Q;i++)
 87         {
 88             int x,y;
 89             scanf("%d%d",&x,&y);
 90             flag[x-1][y-1]=1;
 91             flag[y-1][x-1]=1;
 92         }
 93         for (int i=0;i<n;i++)
 94         {
 95             for (int j=0;j<n;j++)
 96             {
 97                 if (i!=j)
 98                 {
 99                     if (flag[i][j]==1)
100                     {
101                         addnode(i,j,0);    
102                     }
103                     else
104                     {
105                         addnode(i,j,a[i][j]);
106                     }
107                 }
108             }
109         }
110         kruskal();
111         printf("%d\n",ans);
112     }
113     return 0;
114 }