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2 years ago
#include <bits/stdc++.h>
const int N = 100010, M = 2 * N;
const int INF = 0x3f3f3f3f;
using namespace std;
int n, m; // 总点数
int d[N]; // 存储每个点到1号点的最短距离
int st[N]; // 存储每个点是否在队列中
// 邻接表
int e[M], h[N], idx, w[M], ne[M];
void add(int a, int b, int c) {
e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
}
// 求1号点到n号点的最短路距离如果从1号点无法走到n号点则返回-1
void spfa(int s) {
// 初始化距离
memset(d, 0x3f, sizeof d);
d[s] = 0;
queue<int> q;
q.push(s);
st[s] = 1;
while (q.size()) {
int u = q.front();
q.pop();
st[u] = 0;
for (int i = h[u]; ~i; i = ne[i]) {
int v = e[i];
if (d[v] > d[u] + w[i]) {
d[v] = d[u] + w[i];
// 如果队列中已存在v则不需要将v重复插入,优化一下
if (!st[v]) {
q.push(v);
st[v] = 1;
}
}
}
}
}
int main() {
memset(h, -1, sizeof h);
cin >> n >> m;
while (m--) {
int a, b, c;
cin >> a >> b >> c;
add(a, b, c);
}
spfa(1);
if (d[n] == INF)
puts("impossible");
else
printf("%d\n", d[n]);
return 0;
}