[暑假集训Day2T3]团建活动

个人认为这周题中较难的一道。

题意大概为：给定一张N个点M条边的无向图，求出无向图的一棵最小生成树，满足一号节点的度数不超过给定的整数K。保证 N <= 20

首先用map存取节点，之后抛去1号节点，求每一个联通分量的MST，就得到了一个局部最优解，设p为联通块的个数，接下来从每一个联通分量中找一个离1号节点距离最近的点，连接其与1号节点的边，就得到了一棵1号节点度数为p的生成树，之后逐渐放宽限制，枚举最终答案的度数，每次从1号节点往其他点连边，会形成一个环，求出该点到1路径上形成的环中边权最大值，之后把这个最大值剪断，将1号节点和该点连边。直到减小不了权值为止

下面给出标程（Author:????）

#include <map>
#include <cstdio>
#include <string>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 37, INF = 0x3f3f3f3f;
struct E {
    int x, y, z;
    bool operator < (const E w) const
    {
        return z < w.z;
        }    
         
    }f[N];
int n, k, tot, ans, a[N][N], fa[N], d[N], v[N];
map<string, int> m;
vector<E> e;
bool b[N][N];
 
int get(int x) {
    if (x == fa[x]) return x;
    return fa[x] = get(fa[x]);
}
// dfs(1,-1)
void dfs(int x, int o) {
    for (int i = 2; i <= tot; i++) {
        if (i == o || b[x][i]==0) continue;
        if (f[i].z == -1) {
            if (f[x].z > a[x][i]) f[i] = f[x];
            else {
                f[i].x = x;
                f[i].y = i;
                f[i].z = a[x][i];
            }
        }
        dfs(i, x);
    }
}
 
int main() {
    cin >> n;
    memset(a, 0x3f, sizeof(a));
    memset(d, 0x3f, sizeof(d));
    m["Park"] = tot = 1;
    for (int i = 0; i < N; i++) fa[i] = i;
    for (int i = 1; i <= n; i++) {
        E w;
        string s1, s2;
        cin >> s1 >> s2 >> w.z;
        w.x = m[s1] ? m[s1] : (m[s1] = ++tot);
        w.y = m[s2] ? m[s2] : (m[s2] = ++tot);
        e.push_back(w);
        a[w.x][w.y] = a[w.y][w.x] = w.z;
    }
    cin >> k;
    sort(e.begin(), e.end());
    for (unsigned int i = 0; i < e.size(); i++) {
        if (e[i].x == 1 || e[i].y == 1) continue;
        int rtx = get(e[i].x), rty = get(e[i].y);
        if (rtx != rty) {
            fa[rtx] = rty;
            b[e[i].x][e[i].y] = b[e[i].y][e[i].x] = 1;
            ans += e[i].z;
        }
    }
    for (int i = 2; i <= tot; i++)
        if (a[1][i] != INF) {
            int rt = get(i);
            if (d[rt] > a[1][i]) 
                d[rt] = a[1][v[rt]=i]; // 记下每个父节点对应的边缘点以及边缘点和1号点的权值 
        }
    for (int i = 1; i <= tot; i++)
        if (d[i] != INF) {
            --k;
            b[1][v[i]] = b[v[i]][1] = 1;
            ans += a[1][v[i]];
        }
        //cout<<k<<endl;
    while (k--) {
        memset(f, -1, sizeof(f));
        f[1].z = -INF;
        for (int i = 2; i <= tot; i++)
            if (b[1][i]) f[i].z = -INF;
        dfs(1, -1);
        int o, w = -INF;
        for (int i = 2; i <= tot; i++)
            if (w < f[i].z - a[1][i])
                w = f[i].z - a[1][o=i];
        if (w <= 0) break;
        //cout<<o<<endl;
        b[1][o] = b[o][1] = 1;
        b[f[o].x][f[o].y] = b[f[o].y][f[o].x] = 0;
        ans -= w;
    }
    printf("Total miles driven: %d
", ans);
    return 0;
}