例题
- luogu P3366 【模板】最小生成树 (以下代码均可A此题)
kruskal
-
Kruskal算法通过并差集维护,从到小枚举每条边,如果两端点不在一个集合,将两端点所在集合合并,并将边权累加到答案中
-
时间复杂度为(O(m log m))
#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 5005, M = 2e5+5;
struct side {
int x, y, d;
}e[M];
bool operator < (side a, side b) {
return a.d < b.d;
}
int n, m, f[N], ans, cnt;
int found(int x) {
return x == f[x] ? x : (f[x] = found(f[x]));
}
void kruskal() {
for (int i = 1; i <= m; i++) {
int x = found(e[i].x), y = found(e[i].y);
if (x == y) continue;
f[x] = y;
ans += e[i].d;
if (++cnt == n-1) return;//最小生成树上最多n-1条边
}
}
int main() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= m; i++)
scanf("%d%d%d", &e[i].x, &e[i].y, &e[i].d);
sort(e + 1, e + m + 1);//排序
for (int i = 1; i <= n; i++) f[i] = i;//并查集初始化
kruskal();
if (cnt == n-1) printf("%d
", ans);
else puts("orn");//奇奇怪怪的输出,其实他没有这个测试点
return 0;
}
prim
-
Prim算法思想类似于Dijkatra。
维护一个数组 d[x] 节点 x 与已确定是最小生成树集合中的节点之间权值最小的边的权值
每次从未标记的节点中选出 d 值最小的点,将其标记,在更新其他点的 d 值 -
时间复杂度为(O(n^2)),使用堆优化可以到(O(mlogn))(这里写的是堆优化后的板子) 但是这样不如Kruskal方便,
因此,Prim主要用于稠密图,尤其是完全图的求解
#include <queue>
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 5005, M = 2e5+5;
struct side{
int t, d, next;
}e[M<<1];
int head[N], tot;
void add(int x, int y, int z) {
e[++tot].next = head[x];
head[x] = tot;
e[tot].t = y;
e[tot].d = z;
}
priority_queue< pair<int, int> > q;
int n, m, d[N], ans, cnt;
bool v[N];
void prim() {
memset(d, 0x3f, sizeof(d));
q.push(make_pair(d[1] = 0, 1));
while (!q.empty() && cnt < n) {
int x, w;
x = q.top().second;
w = -q.top().first;
q.pop();
if (v[x]) continue;
v[x] = 1;
ans += w;
cnt++;
for (int i = head[x]; i; i = e[i].next) {
int y = e[i].t;
if (d[y] > e[i].d) {
d[y] = e[i].d;
q.push(make_pair(-d[y], y));
}
}
}
}
int main() {
scanf("%d%d", &n, &m);
while (m--) {
int x, y, z;
scanf("%d%d%d", &x, &y, &z);
add(x, y, z); add(y, x, z);
}
prim();
if (cnt == n) printf("%d
", ans);
else puts("orn");
return 0;
}
Boruvka
-
思路是对于每个联通块找一条连向其他联通块的边权最小的边,然后合并这两个联通块
-
每次操作都会至少减少一半的联通块,每次需要o(m)得遍历每条边,最多便利logn次,时间复杂度 (O(m log n))
-
复杂度瓶颈在于如何找联通块之外边权最小的边,在一些有特殊性质的题中可以用数据结构或性质进行优化
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 2e5 + 5;;
int read(int x = 0, int f = 1, char c = getchar()) {
for (; c < '0' || c > '9'; c = getchar()) if (c == '-') f = -1;
for (; c >='0' && c <='9'; c = getchar()) x = x * 10 + c - '0';
return x * f;
}
int n, m, p[N], w[N], f[N], cnt, ans;
struct Side {
int x, y, d;
}a[N];
int Find(int x) {
return x == f[x] ? x : f[x] = Find(f[x]);
}
int main() {
n = read(); m = read();
for (int i = 1; i <= m; ++i)
a[i] = (Side) {read(), read(), read()};
for (int i = 1; i <= n; ++i) f[i] = i;
while (cnt < n - 1) {
memset(p + 1, 0, n * 4);
memset(w + 1, 0x3f, n * 4);
for (int i = 1; i <= m; ++i) {
int x = Find(a[i].x), y = Find(a[i].y), d = a[i].d;
if (x == y) continue;
if (d < w[x]) w[x] = d, p[x] = y;
if (d < w[y]) w[y] = d, p[y] = x;
}
for (int i = 1; i <= n; ++i) {
if (i != Find(i)) continue;
int x = Find(p[i]);
if (i == x) continue;
f[i] = x, ans += w[i], cnt++;
}
}
printf("%d
", ans);
return 0;
}