POJ2421 Constructing Roads
阿新 • • 發佈：2017-05-07
print cti tdi pla 一個 def clu fine n) 
Constructing Roads
這道題很水，就是一個裸的最小生成樹，最不過把已經連接的節點的值再設為0。
代碼：
1 #include<cstdio> 2 #include<cstring> 3 #define N 1010 4 int dis[N][N],minn[N],n,m,a,b; 5 bool u[N]; 6 int main(){ 7 scanf("%d",&n); 8 for(int i=1;i<=n;++i) 9 for(int j=1;j<=n;++j) 10 scanf("View Code%d",&dis[i][j]); 11 scanf("%d",&m); 12 for(int i=1;i<=m;++i){ 13 scanf("%d%d",&a,&b); 14 dis[a][b]=0; 15 dis[b][a]=0; 16 } 17 memset(minn,0x7f,sizeof(minn)); 18 minn[1]=0; 19 memset(u,1,sizeof(u)); 20 for(int i=1;i<=n;++i){ 21 intk=0; 22 for(int j=1;j<=n;++j) 23 if(u[j]&&(minn[j]<minn[k])) 24 k=j; 25 u[k]=0; 26 for(int j=1;j<=n;j++) 27 if(u[j]&&(dis[k][j]<minn[j])) 28 minn[j]=dis[k][j]; 29 } 30 int total=0; 31 for(int i=1;i<=n;++i) 32 total+=minn[i]; 33 printf("%d",total); 34 return 0; 35 }
POJ2421 Constructing Roads