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 = 500010;
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 = 0; i < m; ++i)
        scanf("%d%d%d%d",&e[i].u,&e[i].v,&e[i].length,&e[i].cost);
        for(int i = 0; i <= n; ++i) uf[i] = i;
        sort(e,e + m);
        LL length = 0,cost = 0,cnt = 0;
        for(int i = 0; i < m && cnt + 1 < 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 + 1 == n) printf("%lld %lld
",length,cost);
        else puts("-1 -1");
    }
    return 0;
}