[BZOJ1083][SCOI2005]繁忙的都市 题解

BZOJ:https://www.lydsy.com/JudgeOnline/problem.php?id=1083

这题就是一道最小生成树的裸题

我使用的是 $ kruskal $ 算法。

题目的第一问就是 $ n - 1 $ ，这个是很显然的。

第二问就跑一下 $ kruskal $ 就行了。

$ code $

# include <cstdio>
# include <algorithm>
# include <cstring>
# include <cmath>
# include <string>
# define ll long long
# define rg register
# define il inline
using namespace std;

const int maxM = 300 * 300 + 10; 
struct Edge
{
    int frm;
    int to;
    int val;
}edge[maxM];

il void add_edge(int, int, int);
il int kr();
il bool cmp(Edge, Edge);
int find(int);
void join(int, int);
int n, m, tot = 0, pre[maxM];

int main()
{
    scanf("%d%d", &n, &m);
    for(rg int i = 1; i <= n; ++ i)
        pre[i] = i; 
    for(rg int i = 0; i < m; ++ i)
    {
        rg int x, y, z;
        scanf("%d%d%d", &x, &y, &z);
        add_edge(x, y, z);
    }
    sort(edge, edge + tot, cmp);
    printf("%d %d", n - 1, kr());
}

il void add_edge(int x, int y, int z)
{
    edge[tot].frm = x;
    edge[tot].to = y;
    edge[tot ++].val = z;
}

il bool cmp(Edge a, Edge b)
{
    return a.val < b.val;
}

il int kr()
{
    rg int ans;
    for(rg int i = 0; i < m; ++ i)
    {
        rg int u = find(edge[i].frm);
        rg int v = find(edge[i].to);
        if(u != v)
        {
            join(u, v);
            ans = edge[i].val;
        }
    }
    return ans;
}

int find(int x)
{
    if(x != pre[x]) pre[x] = find(pre[x]);
    return pre[x];
}

void join(int x, int y)
{
    x = find(x), y = find(y);
    pre[x] = y;
}