67 lines
1.8 KiB
C++
67 lines
1.8 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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const int N = 1e5 + 10, M = N << 1;
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typedef pair<int, int> PII;
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int n, m, s;
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int dist[2 * N];
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// 链式前向星，因为是两份图，所以啥啥都是两份
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int h[2 * N], ne[2 * M], e[2 * M], w[2 * M], idx;
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bool st[N];
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void add(int a, int b, int c) {
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e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
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}
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void dijkstra(int s) {
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memset(dist, 0x3f, sizeof dist);
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dist[s] = 0;
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priority_queue<PII, vector<PII>, greater<PII>> q; // 小顶堆
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q.push({0, s});
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while (q.size()) {
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auto t = q.top();
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q.pop();
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int u = t.second;
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if (st[u]) continue;
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st[u] = 1;
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for (int i = h[u]; ~i; i = ne[i]) {
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int j = e[i];
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if (dist[j] > dist[u] + w[i]) {
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dist[j] = dist[u] + w[i];
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q.push({dist[j], j});
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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); // 正图
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add(b + n, a + n, c); // 反图
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}
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int res = 0;
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// 单源最短路，就是从1出发，到达其它n-1个点的最短距离
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dijkstra(1);
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for (int i = 1; i <= n; i++) res += dist[i];
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// 从n-1个节点，返回1号节点，就是多源最短路
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// 通过建反图，将多源最短路转化为单源最短路
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// 建反图时，所以节点号+n处理，这样1号节点就变成1+n号节点,其它n-1个节点向1号节点的多源最短路就变成了
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// 从1+n~ [2+n,3+n,...,n+n]的最短路径了
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dijkstra(1 + n);
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for (int i = 1 + n; i <= 2 * n; i++) res += dist[i];
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printf("%d\n", res);
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return 0;
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}
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