最小k度最小生成树模板

代码是抄的

题解是瞄的

可我想学习的心是真的嘤嘤嘤

然而

还是上传一份ioi大神的论文吧

链接：https://pan.baidu.com/s/1neIW9QeZEa0hXsUqJTjmeQ

密码：blr4

代码如下

#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <stack>
#include <queue>
#include <cstdio>
#include <cctype>
#include <bitset>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
#define fuck(x) cout<<"["<<x<<"]";
#define FIN freopen("input.txt","r",stdin);
#define FOUT freopen("output.txt","w+",stdout);
//#pragma comment(linker, "/STACK:102400000,102400000")
 
const int INF = 0x3f3f3f3f;
const int maxn = ;
 
using namespace std;
 
struct Edge{
  int u, v, d;
  Edge() {}
  Edge(int a, int b, int c) : u(a), v(b), d(c) {}
  bool operator < (const Edge &e) const {
    return d < e.d;
  }
};
 
int n, m, k;
int cnt;
int ans;
int f[maxn]; // 并查集
map<string, int> nodes;
vector<Edge> edges;
Edge dp[maxn];
int g[maxn][maxn];
bool tree[maxn][maxn]; // tree[i][j]=true表示<i, j>这条边在最小生成树中
int minEdge[maxn];
 
int find(int p) {
  if (p == f[p]) return f[p];
  return f[p] = find(f[p]);
}
 
void unite(int p, int q) {
  f[find(p)] = find(q);
}
 
void kruskal() {
  sort(edges.begin(), edges.end());
  for (int i = ; i < edges.size(); i++) {
    int p = edges[i].u;
    int q = edges[i].v;
    if (p ==  || q == ) continue; // 忽略根节点
    if (find(p) != find(q)) {
      unite(p, q);
      tree[p][q] = tree[q][p] = ;
      ans += edges[i].d;
    }
  }
}
 
void dfs(int cur, int pre) {
  for (int i = ; i <= cnt; i++) {
    if (i == pre || !tree[cur][i]) continue;
    if (dp[i].d == -) {
      if (dp[cur].d > g[cur][i]) dp[i] = dp[cur];
      else {
        dp[i].u = cur;
        dp[i].v = i;
        dp[i].d = g[cur][i];
      }
    }
    dfs(i, cur);
  }
}
 
void solve() {
  int keyPoint[maxn];
  for (int i = ; i <= cnt; i++) {
    if (g[][i] != INF) {
      // 点i在哪颗最小生成树中
      int color = find(i);
      // 每颗最小生成树中距离根节点最近的点与根节点的距离
      if (minEdge[color] > g[][i]) {
        minEdge[color] = g[][i];
        keyPoint[color] = i;
      }
    }
  }
  for (int i = ; i <= cnt; i++) {
    if (minEdge[i] != INF) {
      m++;
      tree[][keyPoint[i]] = tree[keyPoint[i]][] = ;
      ans += g[][keyPoint[i]];
    }
  }
  // 由i-1度生成树得i度生成树
  for (int i = m + ; i <= k; i++) {
    memset(dp, -, sizeof(dp));
    dp[].d = -INF;
    for (int j = ; j <= cnt; j++)
      if (tree[][j]) dp[j].d = -INF;
    dfs(, -); // dp预处理
    int idx, minnum = INF;
    for (int j = ; j <= cnt; j++) {
      if (minnum > g[][j] - dp[j].d) {
        minnum = g[][j] - dp[j].d;
        idx = j;
      }
    }
    if (minnum >= ) break;
    tree[][idx] = tree[idx][] = ;
    tree[dp[idx].u][dp[idx].v] = tree[dp[idx].v][dp[idx].u] = ;
    ans += minnum;
  }
}
 
void init() {
  memset(g, 0x3f, sizeof(g));
  memset(tree, , sizeof(tree));
  memset(minEdge, 0x3f, sizeof(minEdge));
  m = ;
  cnt = ;
  ans = ;
  nodes["Park"] = ;
  for (int i = ; i < maxn; i++)
    f[i] = i;
}
 
int main() {
#ifndef ONLINE_JUDGE
  FIN
#endif
  scanf("%d", &n);
  string s1, s2;
  int d;
  init();
  for (int i = ; i <= n; i++) {
    cin >> s1 >> s2 >> d;
    if (!nodes[s1]) nodes[s1] = ++cnt;
    if (!nodes[s2]) nodes[s2] = ++cnt;
    int u = nodes[s1], v = nodes[s2];
    edges.push_back(Edge(u, v, d));
    g[u][v] = g[v][u] = min(g[u][v], d);
  }
  scanf("%d", &k);
  kruskal(); // 忽略根节点先计算一次最小生成树，此时得到一个森林
  solve();
  printf("Total miles driven: %d\n", ans);
  return ;
}