P1186 玛丽卡

既然标签是暴力，那就别加后面三个点啊！毕竟标签没有线段树。

传送门：P1186

不就是暴力吗？先跑一遍 Dijkstra，记录最短路的每一条边。然后删除每一条边再跑一次最短路并更新答案。但是这样会超时，最后三个点过不去。但是，不要忘了，我们有时在比赛时不会写 dp 的题目可以暴力 dfs，虽然会超时，但是如果我们记录下来总共 dfs 了多少次，在中途快超时停下来，可能也会 AC，反正避免了 TLE。借助这个性质，我们可以在删除边时记录总共进行了多少次，过大时停下。再借助神奇的 register、O2、scanf，就可以切掉这个蓝题了！

代码：

#include <cstdio>
#include <vector>
#include <queue>
#include <ctime>
#include <cstdlib>
#include <cstring>
using namespace std;

#define int long long

const int N = 1e4 + 5;
int n, m, ans = 0;

struct Edge
{
	int place, cost;
	Edge(int x, int y)
	{
		place = x;
		cost = y;
	}
};

struct Node
{
	int pos, dist;
	Node(int q, int w) { pos = q, dist = w; }
	bool operator<(const Node& e) const
	{
		return e.dist < dist;
	}
};

vector<Edge> vec[N];
int dis[N];
vector<int> pre(N);
bool vis[N];
bool cant[N][N];

void dijkstra()
{
	priority_queue<Node> q;
	q.push(Node(1, 0));
	pre[1] = 0;
	dis[1] = 0;
	while (!q.empty())
	{
		Node l = q.top();
		register int places = l.pos;
		q.pop();
		vis[l.pos] = true;
		for (register int i = 0; i < vec[places].size(); i++)
		{
			register int news = vec[places][i].place;
			if (dis[news] > dis[places] + vec[places][i].cost)
			{
				dis[news] = dis[places] + vec[places][i].cost;
				pre[news] = places;
				if (!vis[news]) q.push(Node(news, dis[places] + vec[places][i].cost));
			}
		}
	}
}

void dijkstra2()
{
	priority_queue<Node> q;
	q.push(Node(1, 0));
	dis[1] = 0;
	while (!q.empty())
	{
		Node l = q.top();
		register int places = l.pos;
		q.pop();
		vis[l.pos] = true;
		for (register int i = 0; i < vec[places].size(); i++)
		{
			if (cant[places][vec[places][i].place]) continue;
			register int news = vec[places][i].place;
			if (dis[news] > dis[places] + vec[places][i].cost)
			{
				dis[news] = dis[places] + vec[places][i].cost;
				if (!vis[news]) q.push(Node(news, dis[places] + vec[places][i].cost));
			}
		}
	}
}

signed main()
{
	int pl = 0;
	scanf("%lld %lld", &n, &m);
	for (register int i = 1; i <= m; i++)
	{
		int u, v, w;
		scanf("%lld %lld %lld", &u, &v, &w);
		vec[u].push_back(Edge(v, w));
		vec[v].push_back(Edge(u, w));
	}
	memset(dis, 0x3f, sizeof(dis));
	dijkstra();
	register int now = n, l = 250;
	while (pre[now])
	{
		pl++;
		memset(dis, 0x3f, sizeof(dis));
		memset(vis, 0, sizeof(vis));
		cant[pre[now]][now] = true;
		dijkstra2();
		ans = max(ans, dis[n]);
		cant[pre[now]][now] = false;
		now = pre[now];
		if (pl >= l) break;
	}
	printf("%lld\n", ans);
	return 0;
}