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
", dis2[1]);
return 0;
}