Light OJ 1041

题目链接

http://www.lightoj.com/volume_showproblem.php?problem=1041

题目大意

有m个关系， 每个关系是两个城市相距的距离，求把这些城市全部连接起来的最小距离。

解题思路

典型的最小生成树

代码如下

kruskal算法

#include<bits/stdc++.h>
using namespace std;
 
const int N = 57;
const int INF = 0x3f3f3f3f;
 
struct edge
{
    int u, v, cost;
    edge() = default;
    edge(int x, int y, int z)
    {
        u = x, v = y, cost = z;
    }
};
edge arr[N];
 
 
map<string, int>mp;
set<string> st;
 
int par[N];
int Rank[N];
 
 
void init(int n)
{
    for(int i=1; i<=n; ++ i)
        par[i] = i, Rank[i] = 0;
}
 
int Find(int x)
{
    if(x == par[x])
        return x;
    return par[x] = Find(par[x]);
}
 
void unite(int x, int y)
{
    x = Find(x);
    y = Find(y);
    if(x == y)
        return ;
    if(Rank[x] < Rank[y])
        par[x] = y;
    else
    {
        par[y] = x;
        if(Rank[x] == Rank[y])
            Rank[x] ++;
    }
}
 
bool same(int x, int y)
{
    if(Find(x) == Find(y))
        return true;
    return false;
}
bool cmp(const edge &a, const edge &b)
{
    return a.cost < b.cost;
}
 
int kruskal(int e, int v)
{
    sort(arr, arr+e, cmp);
    init(v);
    int res = 0;
    for(int i=0; i<e; ++ i)
    {
        edge s = arr[i];
        if(same(s.u, s.v) == false)
        {
            unite(s.u, s.v);
            res += s.cost;
        }
    }
    return res;
}
 
void solve(int cases)
{
    int n;
    scanf("%d", &n);
 
    string a, b;
    int c;
 
    int s = 1;
    mp.clear(), st.clear();
    for(int i=0; i<n; ++ i)
    {
        cin >> a >> b >> c;
        if(st.count(a) == 0)
        {
            st.insert(a);
            mp[a] = s ++;
        }
        if(st.count(b) == 0)
        {
            st.insert(b);
            mp[b] = s ++;
        }
        arr[i] = edge(mp[a], mp[b], c);
    }
 
    int res = kruskal(n, s-1);
    int is = 0;
    for(int i=1; i<s; ++ i)
        if(par[i] == i)
        is ++;
    if(is > 1)
        printf("Case %d: Impossible
", cases);
    else
        printf("Case %d: %d
", cases, res);
}
 
int main()
{
    int t;
    scanf("%d", &t);
    for(int i=1; i<=t; ++ i)
        solve(i);
    return 0;
}