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