樹狀結構學習（1）、最大-最小劃分樹

樹狀結構學習（1）、最大-最小劃分樹

有這樣一類題：
給一棵n個點的樹，要求將它分成k個子樹，使權值和最小的子樹權值和儘量大。
解法一：二分答案
我們知道，如果已知權值下限，我們可以從下往上搜以O（n）的複雜度判斷是否可以分成k個權值大於權值下限的子樹。
缺點：沒有真正解出這道題，如果權值是實數或很大將會使這種方法失效。

#include <iostream>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
const int MAXN = 1e5 + 10;

int n, k;
int size[MAXN];
int bri[MAXN];
int fa[MAXN];
int w[MAXN];
vector<int> map[MAXN];

void dfs(int x, int father) {
    size[x] = w[x];
    for (int i = 0; i < map[x].size(); ++i) {
      int y = map[x][i];
      if (y == father) continue; 
      fa[y] = x;
      dfs(y, x);
      size[x] += size[y];
    }
}

int tot;

void dfs(int x, int father, int val) {
    int delta = w[x];
    for (int i = 0; i < map[x].size(); ++i) {
      int y = map[x][i];
      if (y == father) continue;
      dfs(y, x, val);
      delta += size[y];
    }
    size[x] = delta;
    if (size[x] >= val) {
      ++tot;
      size[x] = 0;
    }
}

bool check(int x) {
    tot = 0;
    for (int i = 1; i <= n; ++i)
      bri[i] = size[i];
    dfs(1, -1, x);
    for (int i = 1; i <= n; ++i)
      size[i] = bri[i];
    if (tot >= k) return true;
    else return false;
}

int main()
{
    freopen("in.txt", "r", stdin);
    freopen("out1.txt", "w", stdout);
    int a, b;
    scanf("%d%d", &n, &k);
    for (int i = 1; i < n; ++i) {
      scanf("%d%d", &a, &b);
      map[a].push_back(b);
      map[b].push_back(a);
    }
    int l = 1, r = 0;
    for (int i = 1; i <= n; ++i)
      scanf("%d", &w[i]), r += w[i];
    r /= k;
    dfs(1, -1);
    while (l < r) {
      int mid = (l + r + 1) >> 1;
      if (check(mid)) 
       `
l = mid;
      else
        r = mid - 1;
    }
    printf("%d", l); 
    return 0;
} 

解法二：
移動割法
算法流程：
（1）選擇入度爲1的結點作爲起始，將k-
1個割都放在與根相連的唯一的邊上。
（2）計算當前劃分的最小值Wmin
（3）移動割，並求出所有移動方法最大的子樹權值和Wmax
（4）如果Wmax > Wmin，那麼移動，並跳到（2）
（5）結束。Wmin爲所求。當前割的位置即爲一個可行的最優方案。

算法證明之後傳。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
const int MAXN = 100000 + 10;
const int INF = 0x7fffffff;

int n, k, Wmin, start; 
int in[MAXN];   //入度 
int size[MAXN];
int fa[MAXN];   //father
int vec[MAXN];  //每一個割現在的容量
int have[MAXN]; //結點x是否有一個割 
int pos[MAXN];  //割的位置 
int w[MAXN];    //權重 
int totSize = 0;
vector<int> map[MAXN];

void dfs(int x, int father) {
    size[x] = w[x];
    totSize += size[x];
    for (int i = 0; i < map[x].size(); ++i) {
      int y = map[x][i];
      if (y == father) continue;
      fa[y] = x;
      dfs(y, x); 
      size[x] += size[y];
    }
}

bool down() {
    int Wmax = 0, next = 0, id = 0;
    for (int i = 1; i <= k; ++i) {
      int x = pos[i];
      for (int j = 0; j < map[x].size(); ++j) {
        int y = map[x][j];
        if (y == fa[x] || have[y]) continue;
        if (size[y] >= Wmin && size[y] > Wmax) {
          Wmax = size[y];
          next = y;
          id = i;
        }
      }
    }
    if (!next) return false;
    --have[pos[id]];
    int tmp = size[pos[id]] - size[next], i = fa[pos[id]], flag = 1;
    while (i && flag) {
      if (have[i]) flag = 0;
      size[i] += tmp;
      i = fa[i];
    }
    size[pos[id]] -= size[next];
    pos[id] = next; 
    ++have[next];
    return true;
} 

int query() {
/*  cout<<"Status of have[]"<<endl;
    for (int i = 1; i <= n; ++i)
      printf("(%d, %d), ", i, have[i]);
    cout<<endl;
    cout<<"Status of size[]"<<endl;
    for (int i = 1; i <= k; ++i)
      cout<<size[pos[i]]<<" ";*/
    int res = INF, sum = 0;
    for (int i = 1; i <= k; ++i)
      if (pos[i] != pos[i-1]) {
        res = min(res, size[pos[i]]);
        sum += size[pos[i]];
      }
//  cout<<totSize - sum<<endl;
    res = min(res, totSize - sum);
    return res;
}

int move(int start) {
    Wmin = w[start];
    while (true) {
      if (!down()) break;
      Wmin = query();
    }
    return Wmin;      
}

int main()
{
    freopen("in.txt", "r", stdin);
    freopen("out2.txt", "w", stdout);
    int a, b, Wstart = INF;
    scanf("%d%d", &n, &k);
    --k;
    for (int i = 1; i < n; ++i) {
      scanf("%d%d", &a, &b);
      map[a].push_back(b);
      map[b].push_back(a);
      ++in[a];
      ++in[b];
    }
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &w[i]);
      if ((in[i] == 0 || in[i] == 1) && w[i] < Wstart)
        start = i, Wstart = w[i];
    }
    dfs(start, -1);
    if (k == 0) printf("%d", totSize);
    else {
        for (int i = 1; i <= k; ++i)
          pos[i] = map[start][0];
        have[map[start][0]] = k;
        printf("%d", move(start));  
    }
    return 0;
}

附：隨機數生成程序

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <ctime>
using namespace std;

int main()
{
    srand(time(0));
    freopen("in.txt", "w", stdout); 
    int n = rand()%10000 + 1, k = rand()%n / 2 + 1;
    cout<<n<<" "<<k<<endl;
    for (int i = 2; i <= n; ++i)
      cout<<rand()%(i-1)+1<<" "<<i<<endl;;
    for (int i = 1; i <= n; ++i)
      cout<<rand()%100000 + 1<<endl;
    return 0;
}