CCF CSP 201703 4.地铁修建

套路题。两个方法，并查集或者SPFA，推荐用并查集。

第一个，并查集

#include <iostream>
#include <algorithm>
#include <vector>
#include <cstring>
using namespace std;

const int MAXN = 1e5 + 2;
int N, M;
int pre[MAXN];

struct Edge
{
	Edge(int _n1, int _n2, int _days) :n1(_n1), n2(_n2), days(_days) {}
	int n1, n2, days;
	bool operator<(const Edge& e) const
	{
		return days < e.days;
	}
};
vector<Edge> ve;

int f(int x)
{
	int f0 = x, f1 = x;
	for (; pre[f0] > 0;) f0 = pre[f0];
	for (; pre[f1] > 0;)
	{
		int t = f1;
		f1 = pre[f1];
		pre[t] = f0;
	}
	return f0;
}

void u(int n1, int n2)
{
	int f1 = f(n1);
	int f2 = f(n2);
	if (f1 != f2)
	{
		if (pre[f1] <= pre[f2])
		{
			pre[f1] += pre[f2];
			pre[f2] = f1;
		}
		else
		{
			pre[f2] += pre[f1];
			pre[f1] = f2;
		}
	}
}

int main()
{
	memset(pre, -1, sizeof pre);
	int a, b, c;
	scanf("%d%d", &N, &M);
	for (int i = 0; i < M; i++)
	{
		scanf("%d%d%d", &a, &b, &c);
		//ve.push_back({ a,b,c });        // 用这个会编译错误的
		ve.push_back(Edge(a, b, c));
	}
	sort(ve.begin(), ve.end());
	for (int i = 0; i < ve.size(); i++)
	{
		u(ve[i].n1, ve[i].n2);
		if (f(1) == f(N))
		{
			printf("%d\n", ve[i].days);
			break;
		}
	}

    return 0;
}

第二个，SPFA

#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
using namespace std;

const int oo = 1e9;
const int MAXN = 1e5 + 5;
int N, M;
bool inq[MAXN] = { 0 };
int opt[MAXN];

struct Edge
{
	Edge(int _from, int _to, int _weight) :from(_from), to(_to), weight(_weight) {}
	int from, to;
	int weight;
};
vector<Edge> ve;
vector<int> v[MAXN];

void bfs(int s)
{
	queue<int> q;
	q.push(s);
	inq[s] = true;
	opt[s] = 0;
	for (; !q.empty();)
	{
		int t = q.front();
		q.pop();
		inq[t] = false;
		for (int i = 0; i < v[t].size(); i++)
		{
			int e = v[t][i];
			int next = ve[e].to;
			int m;
			if ((m = max(opt[t], ve[e].weight)) < opt[next])
			{
				opt[next] = m;
				if (!inq[next])
				{
					q.push(next);
					inq[next] = true;
				}
			}
		}
	}
}

int main()
{
	scanf("%d%d", &N, &M);
	fill(opt + 1, opt + N + 1, oo);
	int f, t, w;
	for (int i = 0; i < M; i++)
	{
		scanf("%d%d%d", &f, &t, &w);
		ve.push_back(Edge(f, t, w));
		ve.push_back(Edge(t, f, w));
		v[f].push_back(i << 1);
		v[t].push_back(i << 1 | 1);
	}
	bfs(1);
	cout << opt[N] << endl;

    return 0;
}

放一个测试用例（样例不行）

6 8
2 3 3
3 6 7
1 4 2
4 5 5
5 6 6
1 2 2
2 4 1
3 5 3

// 6