ACdream 1135 MST

MST

Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)

Problem Description

Given a connected, undirected graph, a spanning tree of that graph is a subgraph that is a tree and connects all the vertices together.  A single graph can have many different spanning trees. We can also assign a weight to each edge, which is a number representing how unfavorable it is, and use this to assign a weight to a spanning tree by computing the sum of the weights of the edges in that spanning tree. A minimum spanning tree (MST) is then a spanning tree with weight less than or equal to the weight of every other spanning tree.
------ From wikipedia
Now we make the problem more complex. We assign each edge two kinds of
weight: length and cost. We call a spanning tree with sum of length less
than or equal to others MST. And we want to find a MST who has minimal
sum of cost.

Input

There are multiple test cases.
The first line contains two integers N and M indicating the number of vertices and edges in the gragh.
The next M lines, each line contains three integers a, b, l and c indicating there are an edge with l length and c cost between a and b.

1 <= N <= 10,000
1 <= M <= 100,000
1 <= a, b <= N
1 <= l, c <= 10,000

Output

For each test case output two integers indicating the sum of length and cost of corresponding MST.
If you can find the corresponding MST, please output "-1 -1".

Sample Input

4 5
1 2 1 1
2 3 1 1
3 4 1 1
1 3 1 2
2 4 1 3

Sample Output

3 3

Source

dut200901102

Manager

dut200901102

 

解题：是的，没错，就是MST，只是是双关键字排序

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long LL;
 const int maxn = ;
 struct arc{
     int u,v,length,cost;
     bool operator<(const arc &rhs)const{
         if(length == rhs.length) return cost < rhs.cost;
         return length < rhs.length;
     }
 }e[maxn];
 int uf[maxn];
 int Find(int x){
     if(x != uf[x]) uf[x] = Find(uf[x]);
     return uf[x];
 }
 int main(){
     int n,m;
     while(~scanf("%d%d",&n,&m)){
         for(int i = ; i < m; ++i)
         scanf("%d%d%d%d",&e[i].u,&e[i].v,&e[i].length,&e[i].cost);
         for(int i = ; i <= n; ++i) uf[i] = i;
         sort(e,e + m);
         LL length = ,cost = ,cnt = ;
         for(int i = ; i < m && cnt +  < n; ++i){
             int u = Find(e[i].u);
             int v = Find(e[i].v);
             if(u == v) continue;
             uf[u] = v;
             length += e[i].length;
             cost += e[i].cost;
             ++cnt;
         }
         if(cnt +  == n) printf("%lld %lld\n",length,cost);
         else puts("-1 -1");
     }
     return ;
 }