【UVA 10600】 ACM Contest and Blackout（最小生成树和次小生成树）

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【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

 #include<cstdio>

 #include<cstdlib>

 #include<cstring>

 #include<iostream>

 #include<algorithm>

 #include<queue>

 using namespace std;

 #define Maxn 110

 #define INF 0xfffffff

 struct node

 {

     int x,y,c,next;

     bool p;

 }tt[Maxn*Maxn],t[Maxn];

 int len;

 int fa[Maxn],first[Maxn];

 bool cmp(node x,node y) {return x.c<y.c;}

 int mymax(int x,int y) {return x>y?x:y;}

 int mymin(int x,int y) {return x<y?x:y;}

 int ffa(int x)

 {

     if(fa[x]!=x) fa[x]=ffa(fa[x]);

     return fa[x];

 }

 void ins(int x,int y,int c)

 {

     t[++len].x=x;t[len].y=y;t[len].c=c;

     t[len].next=first[x];first[x]=len;

 }

 int n,m,mc[Maxn][Maxn];

 bool np[Maxn];

 void dfs(int x,int f,int l)

 {

     for(int i=;i<=n;i++) if(np[i])

         mc[x][i]=mc[i][x]=mymax(mc[f][i],l);

     np[x]=;

     for(int i=first[x];i;i=t[i].next) if(t[i].y!=f)

     {

         dfs(t[i].y,x,t[i].c);

     }

 }

 int main()

 {

     int T;

     scanf("%d",&T);

     while(T--)

     {

         scanf("%d%d",&n,&m);

         for(int i=;i<=m;i++)

         {

             scanf("%d%d%d",&tt[i].x,&tt[i].y,&tt[i].c);

             tt[i].p=;

         }

         sort(tt+,tt++m,cmp);

         for(int i=;i<=n;i++) fa[i]=i;

         int cnt=,a1=,a2=INF;

         len=;

         memset(first,,sizeof(first));

         for(int i=;i<=m;i++)

         {

             if(ffa(tt[i].x)!=ffa(tt[i].y))

             {

                 fa[ffa(tt[i].x)]=ffa(tt[i].y);

                 tt[i].p=;

                 cnt++;

                 ins(tt[i].x,tt[i].y,tt[i].c);

                 ins(tt[i].y,tt[i].x,tt[i].c);

                 a1+=tt[i].c;

             }

             if(cnt==n-) break;

         }

         memset(mc,,sizeof(mc));

         memset(np,,sizeof(np));

         dfs(,,);

         for(int i=;i<=m;i++) if(!tt[i].p&&tt[i].x!=tt[i].y)

         {

             a2=mymin(a2,a1-mc[tt[i].x][tt[i].y]+tt[i].c);

         }

         printf("%d %d\n",a1,a2);

     }

     return ;

 }