acwing算法提高之数据结构--并查集

作者：笔触狂放9 | 2024-04-26 17:11:54

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1 介绍

本专题用来记录并查集相关的题目。

并查集模板：

//初始化
for (int i = 1; i <= n; ++i) { //n为结点数目
	p[i] = i;
}

//查找
find(int x) {
	if (p[x] != x) p[x] = find(p[x]);
	return p[x];
}

//合并
int pa = find(a);
int pb = find(b);
if (pa != pb) {
	p[pa] = pb;
}
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2 训练

题目1：1250格子游戏

C++代码如下，

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 210;
int n, m;
int p[N * N];

int find(int x) {
    if (p[x] != x) p[x] = find(p[x]);
    return p[x];
}

int main() {
    cin >> n >> m;
    for (int i = 0; i < n * n; ++i) p[i] = i;
    
    for (int i = 1; i <= m; ++i) {
        int x, y;
        char op;
        cin >> x >> y >> op;
        int a, b;
        if (op == 'D') {
            a = (x - 1) * n + y - 1;
            b = x * n + y - 1;
        } else {
            a = (x - 1) * n + y - 1;
            b = (x - 1) * n + y;
        }
        int pa = find(a);
        int pb = find(b);
        if (pa == pb) {
            cout << i << endl;
            return 0;
        } else {
            p[pa] = pb;
        }
    }
    
    cout << "draw" << endl;
    return 0;
}
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题目2：1252搭配购买

C++代码如下，

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>

using namespace std;

const int N = 10010;
int n, m, W;
int c[N], w[N];
int p[N];
int sc[N]; //每个连通块的总价格，只针对根结点
int sw[N]; //每个连通块的总价值
int f[N];

int find(int x) {
    if (p[x] != x) p[x] = find(p[x]);
    return p[x];
}

int main() {
    cin >> n >> m >> W;
    for (int i = 1; i <= n; ++i) {
        cin >> c[i] >> w[i];
    }
    
    for (int i = 1; i <= n; ++i) {
        p[i] = i;
        sc[i] = c[i];
        sw[i] = w[i];
    }
    
    for (int i = 1; i <= m; ++i) {
        int a, b;
        cin >> a >> b;
        int pa = find(a);
        int pb = find(b);
        if (pa != pb) {
            p[pa] = pb;
            sc[pb] += sc[pa];
            sw[pb] += sw[pa];
        }
    }
    
    //01背包问题
    int idx = 0;
    for (int i = 1; i <= n; ++i) {
        if (i == p[i]) {
            sc[idx] = sc[i];
            sw[idx] = sw[i];
            idx++;
        }
    }
    
    for (int i = 0; i < idx; ++i) {
        for (int j = W; j >= sc[i]; --j) {
            f[j] = max(f[j], f[j - sc[i]] + sw[i]);
        }
    }
    
    cout << f[W] << endl;
    
    return 0;   
}
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题目3：237程序自动分析

C++代码如下，

#include <iostream>
#include <cstring>
#include <algorithm> 
#include <unordered_map>

using namespace std;

typedef pair<int, int> PII;

const int N = 200010, M = 1e9 + 10;
int p[N];

int find(int x) {
    if (p[x] != x) p[x] = find(p[x]);
    return p[x];
}

int main() {
    int T;
    cin >> T;
    while (T--) {
        memset(p, 0, sizeof p);
        
        unordered_map<int, int> map_b_s;

        int n;
        cin >> n;
        
        vector<PII> equals, not_equals;
        int idx = 0;
        for (int i = 0; i < n; ++i) {
            int a, b, c;
            cin >> a >> b >> c;
            if (c == 0) {
                not_equals.emplace_back(a,b);
            } else {
                equals.emplace_back(a,b);
            }
            
            if (map_b_s.count(a) == 0) {
                map_b_s[a] = idx++;
            }
            
            if (map_b_s.count(b) == 0) {
                map_b_s[b] = idx++;
            }
        }
        
        for (int i = 0; i < idx; ++i) p[i] = i;
        
        for (auto [a, b] : equals) {
            a = map_b_s[a];
            b = map_b_s[b];
            int pa = find(a);
            int pb = find(b);
            if (pa != pb) {
                p[pa] = pb;
            }
        }
        
        bool flag = true;
        for (auto [a, b] : not_equals) {
            a = map_b_s[a];
            b = map_b_s[b];
            int pa = find(a);
            int pb = find(b);
            if (pa == pb) {
                flag = false;
                cout << "NO" << endl;
                break;
            }
        }
        if (flag) {
            cout << "YES" << endl;
        }
    }
    return 0;
}
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题目4：239奇偶游戏

C++代码如下，

//带边权写法
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>

using namespace std;

const int N = 20010;

int n, m;
int p[N], d[N];
unordered_map<int, int> S;

int get(int x) {
    if (S.count(x) == 0) S[x] = ++n;
    return S[x];
}

int find(int x) {
    if (p[x] != x) {
        int root = find(p[x]);
        d[x] += d[p[x]];
        p[x] = root;
    }
    return p[x];
}

int main() {
    cin >> n >> m;
    n = 0;
    
    for (int i = 0; i < N; ++i) p[i] = i;
    
    int res = m;
    for (int i = 1; i <= m; ++i) {
        int a, b;
        string type;
        cin >> a >> b >> type;
        a = get(a - 1), b = get(b);
        
        int t = 0;
        if (type == "odd") t = 1;
        
        int pa = find(a), pb = find(b);
        if (pa == pb) {
            if (((d[a] + d[b]) % 2 + 2) % 2 != t) {
                res = i - 1;
                break;
            }
        } else {
            p[pa] = pb;
            d[pa] = d[a] ^ d[b] ^ t;
        }
    }
    
    cout << res << endl;
    return 0;
}
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题目5：238银河英雄传说

C++代码如下，

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 30010;
int n;
int p[N];
int d[N]; //到根结点的距离
int cnt[N];

int find(int x) {
    if (p[x] != x) {
        int root = find(p[x]);
        d[x] += d[p[x]];
        p[x] = root;
    }
    return p[x];
}

int main() {
    cin >> n;
    
    for (int i = 1; i < N; ++i) {
        p[i] = i;
        cnt[i] = 1;
        d[i] = 0;
    }
    
    for (int i = 0; i < n; ++i) {
        char op;
        int a, b;
        cin >> op >> a >> b;
        int pa = find(a), pb = find(b);
        if (op == 'M') {
            if (pa != pb) {
                d[pa] = cnt[pb];
                cnt[pb] += cnt[pa];
                p[pa] = pb;                
            }
        } else {
            if (pa != pb) puts("-1");
            else {
                cout << max(abs(d[a] - d[b]) - 1, 0) << endl;
            }
        }
    }
    
    return 0;
}
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3 参考

acwing算法基础之数据结构–并查集算法

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acwing算法提高之数据结构--并查集

目录

1 介绍

2 训练

3 参考