[题解] [bzoj2622] 深入虎穴

题解

题解

考虑到正着跑不好想, 我们尝试反向跑

以每个终点作为起点, 维护每个点的最小值和次小值(最小的被老虎ban掉了)

转移的时候用当前点的次小值去更新其所连的点的最小值和次小值

由于最小的次小值不能被其他次小值所更新, 所以我们可以使用dijkstra

把每个终点丢进去跑dijkstra

最后输出\(1\)的次小值即可

Code

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#define itn int
#define reaD read
#define mp(x,y) make_pair(1ll*x,y)
#define N 200005
using namespace std;

int n, m, k, head[N], cnt; 
struct edge { int to, next, cost; } e[N << 5]; 
bool vis[N];
long long dis1[N], dis2[N]; 

namespace Heap
{
	pair<long long, int> heap[N << 2]; int sz = 0; 
	void push(pair <int, int> x) { x.first *= -1; heap[++sz] = x; push_heap(heap + 1, heap + sz + 1); }
	void pop() { pop_heap(heap + 1, heap + sz + 1); sz--; }
	pair<int, int> top() { pair<int, int> tmp; tmp = heap[1]; tmp.first *= -1; return tmp; }
	bool empty() { return !sz; }
}; 

using namespace :: Heap; 

inline int read()
{
	int x = 0, w = 1; char c = getchar();
	while(c < '0' || c > '9') { if (c == '-') w = -1; c = getchar(); }
	while(c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); }
	return x * w;
}

inline void adde(int u, int v, int w) { e[++cnt] = (edge) { v, head[u], w }; head[u] = cnt; }

void dijkstra()
{
	while(!empty())
	{
		int u = top().second; pop();
		if(vis[u]) continue; vis[u] = 1;
		for(int i = head[u]; i; i = e[i].next)
		{
			int v = e[i].to;
			long long sum = dis2[u] + e[i].cost;
			if(sum < dis2[v])
			{
				if(sum < dis1[v]) dis2[v] = dis1[v], dis1[v] = sum;
				else dis2[v] = sum; 
			}
			if(dis2[v] < dis1[0] && !vis[v]) push(mp(dis2[v], v)); 
		}
	}
}

int main()
{
	n = read(); m = read(); k = read();
	for(int i = 1; i <= m; i++)
	{
		itn u = read() + 1, v = read() + 1, w = read();
		adde(u, v, w); adde(v, u, w); 
	}
	memset(dis1, 0x3f, sizeof(dis1)); 
	memset(dis2, 0x3f, sizeof(dis2));
	for(int i = 1; i <= k; i++)
	{
		int x = reaD();
		dis1[x + 1] = dis2[x + 1] = 0;
		push(mp(dis2[x + 1], x + 1)); 
	}
	dijkstra(); 
	printf("%lld\n", dis2[1]); 
	return 0;
}