【题意】
n个点，m条边，求最小生成树的值和次小生成树的值。
Input
The Input starts with the number of test cases, T (1 < T < 15) on a line. Then T test cases follow. The
first line of every test case contains two numbers, which are separated by a space, N (3 < N < 100)
the number of schools in the city, and M the number of possible connections among them. Next M
lines contain three numbers Ai
, Bi
, Ci
, where Ci
is the cost of the connection (1 < Ci < 300) between
schools Ai and Bi
. The schools are numbered with integers in the range 1 to N.
Output
For every test case print only one line of output. This line should contain two numbers separated by a
single space – the cost of two the cheapest connection plans. Let S1 be the cheapest cost and S2 the
next cheapest cost. It’s important, that S1 = S2 if and only if there are two cheapest plans, otherwise
S1 < S2. You can assume that it is always possible to find the costs S1 and S2.
Sample Input
2
5 8
1 3 75
3 4 51
2 4 19
3 2 95
2 5 42
5 4 31
1 2 9
3 5 66
9 14
1 2 4
1 8 8
2 8 11
3 2 8
8 9 7
8 7 1
7 9 6
9 3 2
3 4 7
3 6 4
7 6 2
4 6 14
4 5 9
5 6 10
Sample Output
110 121
37 37
【分析】
主要就是次小生成树咯。
次小生成树一定是最小生成树删一边再加一条边。
先求出最小生成树，然后n^2预处理两点的最小瓶颈路的值maxcost，然后枚举新加入的那条边然后替换边就好了。
啊啊啊啊啊，忘了清bool哭瞎
1 #include<cstdio> 2 #include<cstdlib> 3 #include<cstring> 4 #include<iostream> 5 #include<algorithm> 6 #include<queue> 7 using namespace std; 8 #define Maxn 110 9 #define INF 0xfffffff 10 11 struct node 12 { 13 int x,y,c,next; 14 bool p; 15 }tt[Maxn*Maxn],t[Maxn]; 16 int len; 17 int fa[Maxn],first[Maxn]; 18 19 bool cmp(node x,node y) {return x.c<y.c;} 20 int mymax(int x,int y) {return x>y?x:y;} 21 int mymin(int x,int y) {return x<y?x:y;} 22 23 int ffa(int x) 24 { 25 if(fa[x]!=x) fa[x]=ffa(fa[x]); 26 return fa[x]; 27 } 28 29 void ins(int x,int y,int c) 30 { 31 t[++len].x=x;t[len].y=y;t[len].c=c; 32 t[len].next=first[x];first[x]=len; 33 } 34 35 int n,m,mc[Maxn][Maxn]; 36 bool np[Maxn]; 37 38 void dfs(int x,int f,int l) 39 { 40 for(int i=1;i<=n;i++) if(np[i]) 41 mc[x][i]=mc[i][x]=mymax(mc[f][i],l); 42 np[x]=1; 43 for(int i=first[x];i;i=t[i].next) if(t[i].y!=f) 44 { 45 dfs(t[i].y,x,t[i].c); 46 } 47 } 48 49 int main() 50 { 51 int T; 52 scanf("%d",&T); 53 while(T--) 54 { 55 scanf("%d%d",&n,&m); 56 for(int i=1;i<=m;i++) 57 { 58 scanf("%d%d%d",&tt[i].x,&tt[i].y,&tt[i].c); 59 tt[i].p=0; 60 } 61 sort(tt+1,tt+1+m,cmp); 62 for(int i=1;i<=n;i++) fa[i]=i; 63 int cnt=0,a1=0,a2=INF; 64 len=0; 65 memset(first,0,sizeof(first)); 66 for(int i=1;i<=m;i++) 67 { 68 if(ffa(tt[i].x)!=ffa(tt[i].y)) 69 { 70 fa[ffa(tt[i].x)]=ffa(tt[i].y); 71 tt[i].p=1; 72 cnt++; 73 ins(tt[i].x,tt[i].y,tt[i].c); 74 ins(tt[i].y,tt[i].x,tt[i].c); 75 a1+=tt[i].c; 76 } 77 if(cnt==n-1) break; 78 } 79 memset(mc,0,sizeof(mc)); 80 memset(np,0,sizeof(np)); 81 dfs(1,0,0); 82 for(int i=1;i<=m;i++) if(!tt[i].p&&tt[i].x!=tt[i].y) 83 { 84 a2=mymin(a2,a1-mc[tt[i].x][tt[i].y]+tt[i].c); 85 } 86 printf("%d %d ",a1,a2); 87 } 88 return 0; 89 }
2016-11-01 20:51:55