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ACdream原创群赛(16) F

2024-07-11 17:40:44   221人阅读


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



题意比较坑人，可以存在不连通的图。

这是一个稀疏图，故用kruskal算法！！！




AC代码如下：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#define inf 100000000
using namespace std;
 
int f[10005];
 
struct H{
    int l,r,d,c;
}w[100005];
 
int find (int a)
{
    return f[a]==a? a : f[a]=find(f[a]);
}
 
bool cmp(H a, H b)
{
    return a.d<b.d||(a.d==b.d&&a.c<b.c);//主要注意这个条件的判断就可以了
}
int main()
{
    int n,m;
    int i,j;
    int a,b,c,d;
    int ans;
    while(cin>>n>>m)
    {
        memset(w,0,sizeof w);
        for(i=1;i<=n;i++)
            f[i]=i;
        for(i=1;i<=m;i++)
        {
            cin>>a>>b>>c>>d;
            w[i].l=a;
            w[i].r=b;
            w[i].d=c;
            w[i].c=d;
        }
        int ans=0,cost=0,bj=0;
        sort(w+1,w+m+1,cmp);
        for(i=1;i<=m;i++)
        {
            int xx,yy;
            xx=find(w[i].l);
            yy=find(w[i].r);
            if(xx!=yy)
            {
                bj++;
                ans+=w[i].d;
                cost+=w[i].c;
                f[xx]=yy;
            }
        }
        if(bj==n-1)
        printf("%d %d\n",ans,cost);
        else printf("-1 -1\n");
    }
    return 0;
}


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