lightoj 1029 最小生成树 + 最大生成树
题意，给出若干条连接两个屋子间的路线的价格(保证一定都能联通)，问联通所有屋子的最大代价和最小代价的平均值为多少。
分析，即求一次最大生成树，一次最小生成树
1 /* When all else is lost the future still remains. */ 2 #define rep(X,Y,Z) for(int X=(Y);X<(Z);X++) 3 #define drep(X,Y,Z) for(int X=(Y);X>=(Z);X--) 4 #define fi first 5 #define se second 6 #define mk(X,Y) make_pair((X),(Y)) 7 //head 8 #include <iostream> 9 #include <stdio.h> 10 #include <queue> 11 #include <algorithm> 12 using namespace std; 13 #define MAXN 110 14 pair<int,pair<int,int> > wire[12010]; 15 int rp[MAXN]; 16 void init(){ 17 rep(i,0,MAXN) rp[i] = i; 18 return ; 19 } 20 int f(int x){ 21 return rp[x] = (rp[x] == x) ? x : f(rp[x]); 22 } 23 int MST(int n){ 24 init(); 25 int ans = 0; 26 rep(i,0,n){ 27 int val = wire[i].fi; 28 int a = f(wire[i].se.fi); 29 int b = f(wire[i].se.se); 30 if(a == b) continue; 31 rp[a] = min(a,b); 32 rp[b] = min(a,b); 33 ans += val; 34 } 35 return ans; 36 } 37 int DMST(int n){ 38 init(); 39 int ans = 0; 40 drep(i,n-1,0){ 41 int val = wire[i].fi; 42 int a = f(wire[i].se.fi); 43 int b = f(wire[i].se.se); 44 if(a == b) continue; 45 rp[a] = min(a,b); 46 rp[b] = min(a,b); 47 ans += val; 48 } 49 return ans; 50 } 51 int main(){ 52 int T; 53 scanf("%d",&T); 54 rep(ca,1,T+1){ 55 int num; 56 scanf("%d",&num); 57 int cnt = 0; 58 while(1){ 59 int w , u , v; 60 scanf("%d %d %d",&u,&v,&w); 61 if(!u&&!v&&!w) break; 62 wire[cnt++] = mk(w,mk(u,v)); 63 } 64 sort(wire,wire+cnt); 65 int ansA = MST(cnt); 66 int ansB = DMST(cnt); 67 printf("Case %d: ",ca); 68 if((ansA + ansB) % 2 == 0) printf("%d\n", (ansA + ansB) / 2); 69 else printf("%d/2\n",ansA + ansB); 70 } 71 return 0; 72 }
