题目链接
最小生成树,用prim和kruscal都可以。
(模板题打起来真舒服)
#include <bits/stdc++.h>
using namespace std;
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define inf 0x3f3f3f3f
#define INF 0xffffffff
#define il inline
#define pb push_back
#define db double
#define ll long long
#define ull unsigned long long
#define pii pair<int,int>
#define pll pair<ll,ll>
#define puu pair<ull,ull>
#define MP make_pair
#define fi first
#define se second
const int mod = 1e9+7;
il ll read()
{
char ch = getchar();
ll p = 1,data = 0;
while(ch<'0'||ch>'9')
{
if(ch == '-')p = -1;
ch = getchar();
}
while(ch>='0'&&ch<='9')
{
data = (data<<3)+(data<<1)+(ch^48);
ch = getchar();
}
return p*data;
}
const int maxm = 1e4+5,maxn = 2e3+5;
struct node
{
int from,to,w;
friend bool operator<(node a,node b)
{
return a.w<b.w;
}
} mp[maxm],s[maxm];
int fat[maxn],n,m,cnt,ans,num;
void init()
{
for(int i = 1; i <= n; i++)
fat[i] = i;
}
int find(int x)
{
if(fat[x] == x)
return x;
return fat[x] = find(fat[x]);
}
void unite(int x,int y)
{
int nx = find(x),ny = find(y);
if(nx != ny)
fat[nx] = ny;
}
void kruscal()
{
sort(mp+1,mp+1+num);
for(int i = 1; i <= num; i++)
{
if(find(mp[i].from) != find(mp[i].to))
{
unite(mp[i].from,mp[i].to);
s[++cnt] = mp[i];
}
}
}
int main()
{
n = read(),m = read();
init();
for(int i = 1; i <= m; i++)
{
int p = read(),u = read(),v = read(),w = read();
if(!--p)
{
if(find(u) != find(v))
unite(u,v);
ans += w;
}
else
{
mp[++num].from = u;
mp[num].to = v;
mp[num].w = w;
}
}
kruscal();
for(int i = 1; i <= cnt; i++)
ans += s[i].w;
printf("%d",ans);
return 0;
}
本文深入探讨了最小生成树算法的实现,通过Prim和Kruskal两种方法进行讲解。使用C++代码展示了如何构建最小生成树,并通过实例解释了算法的具体步骤。适合对图论和算法设计感兴趣的学习者。
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