图论━━最短路问题

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

using namespace std;

const int N = 510;

int n, m;
int g[N][N]; // 邻接矩阵
int dist[N]; // 当前顶点1到所有点的距离
bool st[N];  // 是否加入集合S(已知最短距离的点集合)

int dijkstra()
{
	// 初始化当前的距离
	memset(dist, 0x3f, sizeof dist);
	dist[1] = 0;

	for (int i = 0; i < n; i++)
	{
		int t = -1;
		for (int j = 1; j <= n; j++)
			if (!st[j] && (t == -1 || dist[t] > dist[j]))
				t = j;
		st[t] = true;
	
		// 更新相邻顶点的距离
		for (int j = 1; j <= n; j++)
			dist[j] = min(dist[j], dist[t] + g[t][j]);
	}
	
	//判断dist[n]
	if (dist[n] == 0x3f3f3f3f)
		return -1;
	return dist[n];
}

int main()
{
	scanf("%d%d", &n, &m);

	memset(g, 0x3f, sizeof g);
	
	while (m--)
	{
		int a, b, c;
		scanf("%d%d%d", &a, &b, &c);
		// 处理多重边
		g[a][b] = min(g[a][b], c);
	}
	
	int t = dijkstra();
	printf("%d\n", t);
}

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

using namespace std;

using PII = pair<int, int>;
const int N = 100010;
int n, m;
int h[N], w[N], e[N], ne[N], idx;

int dist[N];
bool st[N];

// 邻接表添加边 a --> b (w=c)
void add(int a, int b, int c)
{
	e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}

int dijkstra()
{
	// 初始化距离
	memset(dist, 0x3f, sizeof dist);
	dist[1] = 0;

	priority_queue<PII, vector<PII>, greater<PII>> heap;
	heap.push({0, 1});

	while (heap.size())
	{
		auto t = heap.top();
		heap.pop();

		int ver = t.second, distance = t.first;
		if (st[ver])
			continue;
		// 标记更新过
		st[ver] = true;
		// 只遍历所有的邻边
		for (int i = h[ver]; i != -1; i = ne[i])
		{
			int j = e[i];
			if (dist[j] > distance + w[i])
			{
				dist[j] = distance + w[i];
				heap.push({dist[j], j});
			}
		}
	}

	if (dist[n] == 0x3f3f3f3f)
		return -1;
	return dist[n];
}

int main()
{
	scanf("%d%d", &n, &m);

	// 邻接表的表头初始化
	memset(h, -1, sizeof h);

	while (m--)
	{
		int a, b, c;
		scanf("%d%d%d", &a, &b, &c);
		// 有重边也不影响，算法会使用最短的来更新
		add(a, b, c);
	}

	int t = dijkstra();
	printf("%d\n", t);
	return 0;
}

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

using namespace std;

const int N = 510, M = 10010;

int n, m, k;
int dist[N], backup[N];

struct Edge
{
	int a, b, w;
} edges[M];

int bellman_ford()
{
	memset(dist, 0x3f, sizeof dist);
	dist[1] = 0;

	for (int i = 0; i < k; i++)
	{
		memcpy(backup, dist, sizeof dist);
		for (int j = 0; j < m; j++)
		{
			int a = edges[j].a, b = edges[j].b, w = edges[j].w;
			dist[b] = min(dist[b], dist[a] + w);
		}
	}
	// if (dist[n] > 0x3f3f3f3f / 2)
	// 	return -1;
	// return dist[n];
	// -1也可能是路径长度
	return dist[n];
}

int main()
{
	cin >> n >> m >> k;

	for (int i = 0; i < m; i++)
	{
		int a, b, w;
		cin >> a >> b >> w;
		edges[i] = {a, b, w};
	}

	int res = bellman_ford();
	if (res > 0x3f3f3f3f / 2)
		puts("impossible");
	else
		cout << res << endl;
	return 0;
}

#include <cstring>
#include <algorithm>
#include <queue>

using namespace std;

const int N = 100010;
int n, m;
int h[N], e[N], w[N], ne[N], idx;
int dist[N];
// 优化: 标记是否在队列中
bool st[N];

void add(int a, int b, int c)
{
	e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}

int spfa()
{
	memset(dist, 0x3f, sizeof dist);
	dist[1] = 0;

	queue<int> q;
	q.push(1);
	st[1] = true;
	// 只有更新过的点才能更新其他顶点的距离
	while (!q.empty())
	{
		auto t = q.front();
		q.pop();
		st[t] = false;
		for (int i = h[t]; i != -1; i = ne[i])
		{
			int j = e[i];
			if (dist[j] > dist[t] + w[i])
			{
				dist[j] = dist[t] + w[i];
				if (!st[j])
				{
					q.push(j);
					st[j] = true;
				}
			}
		}
	}
	return dist[n];
}

int main()
{
	scanf("%d%d", &n, &m);

	memset(h, -1, sizeof h);
	for (int i = 0; i < m; i++)
	{
		int x, y, z;
		scanf("%d%d%d", &x, &y, &z);
		add(x, y, z);
	}

	int t = spfa();
	if (t > 0x3f3f3f3f / 2)
		puts("impossible");
	else
		printf("%d\n", t);
	return 0;
}