abc087d <并查集 维护距离信息>

题目

D - People on a Line

思路

• 使用并查集, 维护节点与父节点信息, 检查兼容性

总结

• 注意维护距离信息的并查集的find的写法

int find(int x)
{
    if (x != fa[x])
    {
        int t = fa[x];
        fa[x] = find(fa[x]);
        dis[x] += dis[t];
    }
    return fa[x];
}

• union时, 画图整理distance更新公式, 以防出错

代码

Code

// https://atcoder.jp/contests/abc087/tasks/arc090_b
#include <iostream>
#include <algorithm>
#include <vector>
#include <cstring>

using namespace std;
using LL = long long;
const int N = 1e5 + 10;
int fa[N];
int dis[N];

int find(int x)
{
    if (x != fa[x])
    {
        int t = fa[x];
        fa[x] = find(fa[x]);
        dis[x] += dis[t];
    }
    return fa[x];
}


void solv()
{
    int n, m, l, r, d;
    cin >> n >> m;
    for (int i = 1; i <= n; i ++) fa[i] = i;
    bool flag = true;
    while (m --)
    {
        cin >> l >> r >> d;
        int fl = find(l), fr = find(r);
        if (fl != fr)
        {
            fa[fr] = fl;
            dis[fr] = d + dis[l] - dis[r];
        }
        else if (dis[l] + d != dis[r])
            flag = false;
    }
    cout << (flag? "Yes": "No") << endl;
}

int main()
{
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    int T = 1;
	// cin >> T;
    while (T --)
    {
        solv();
    }
    return 0;
}