AcWing算法提高课【第四章高级数据结构】并查集

1250. 格子游戏 

分析：

显然，当出现闭环的时候，就会出现我们当前节点和要到达的节点出现一个闭环

 代码：

#include <bits/stdc++.h>

using namespace std;

const int N = 210;

int n, m;
int g[N][N], tot;
int f[N * N];

int get(int x)
{
  return f[x] == x ? x : f[x] = get(f[x]);
}
int main()
{
  cin >> n >> m;
  for (int i = 1; i <= n; i ++ ) 
    for (int j = 1; j <= n; j ++ )
      g[i][j] = ++ tot;

  for (int i = 1; i <= n * n; i ++ ) f[i] = i;
  
  for (int i = 1; i <= m; i ++ )
  {
    int a, b; 
    char op[2];
    cin >> a >> b >> op;

    int x = g[a][b], y;
    if (*op == 'D')
        y = g[a + 1][b];
    else 
        y = g[a][b + 1];
    
    x = get(x), y = get(y);
    if (x == y) 
    {
      cout << i << endl;
      return 0;
    }
    f[x] = y;
  }
  
  cout << "draw" << endl;
  
  return 0;
}

1252. 搭配购买 

分析：

显然，彩云被捆绑了。通过并查集捆绑，祖宗节点为fa[x] == x

打包后就是一个0/1背包了。判断是否取当前的祖宗节点。

代码：

#include <cstdio>
#include <cmath>

using namespace std;

const int N = 10010;

int n, m, t;
int v[N], w[N];
int fa[N];
int dp[N];

int get(int x)
{
    return fa[x] == x ? x : fa[x] = get(fa[x]);
}
int main()
{
    scanf("%d%d%d", &n, &m, &t);
    for (int i = 1; i <= n; i ++ )
    {
        fa[i] = i;
        scanf("%d%d", &v[i], &w[i]);
    }
    
    while (m -- ) 
    {
        int x, y; scanf("%d%d", &x, &y);
        x = get(x);
        y = get(y);
        if (x == y) continue;
        fa[x] = y;
        v[y] += v[x];
        w[y] += w[x];
    }
    
    for (int i = 1; i <= n; i ++ )
        if (i == fa[i])
            for (int j = t; j >= v[i]; j -- ) 
                dp[j] = max(dp[j], dp[j - v[i]] + w[i]);
    
    printf("%d
", dp[t]);
    
    return 0;
}