Given a directed graph with n points and m edges, there may be multiple edges and self-loops in the graph, and all edge weights are non-negative.
Please find the shortest distance from point 1 to point n. If it is impossible to go from point 1 to point n, output −1.
Input format
The first line contains integers n and m.
Each of the next m lines contains three integers x, y, z, indicating that there is a directed edge from point x to point y, and the edge length is z.
Output Format
Output an integer representing the shortest distance from point 1 to point n.
If the path does not exist, output −1.
The data range is
1≤n, m≤1.5×105, and
the lengths of the sides involved in the figure are not less than 0 and not more than 10000.
Input sample:
3 3
1 2 2
2 3 1
1 3 4
Output sample:
3
#include<bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
int h[N], e[N], ne[N], w[N], idx;
typedef pair<int, int> PII;
int n, m;
int dist[N];
int vis[N];
void add(int a, int b, int c)
{
e[idx] = b;
ne[idx] = h[a];
w[idx] = c;
h[a] = idx++;
}
int dijkstra()
{
memset(dist, 0x3f, sizeof dist);
dist[1] = 0;
//优先队列模拟堆， greater<> 表示数字越小优先级越大， less< >表示数字越大优先级越大
priority_queue<PII, vector<PII>, greater<PII>> heap;
heap.push({
0, 1});
while(heap.size()) {
auto t = heap.top();
heap.pop();
int v = t.second, d = t.first;
if(vis[v]) continue;
vis[v] = true;
for(int i = h[v]; i!= -1; i = ne[i]) {
int j = e[i];
if(dist[j] > dist[v] + w[i]) {
dist[j] = dist[v] + w[i];
heap.push({
dist[j], j});
}
}
}
if(dist[n] == 0x3f3f3f3f) return -1;
else return dist[n];
}
int main()
{
//初始化
memset(h, -1, sizeof h);
cin >> n >> m;
int a, b, c;
for(int i=0; i<m; i++) {
cin >> a >> b >> c;
add(a, b, c);
}
cout << dijkstra() << endl;
return 0;
}