从 0 到各点的最短距离
七个点的带权图,走不到的记 -1。运行下面这段程序:
#include <algorithm>
#include <iostream>
#include <queue>
#include <set>
#include <string>
#include <tuple>
#include <utility>
#include <vector>
using namespace std;
typedef long long ll;
const ll INF = 1000000000; // 走不到的点记这个值,输出时写成 -1
using Graph = vector<vector<pair<int, int>>>; // g[u] 里放 (邻居 v, 边权 w)
// 七个点的无向图(和 Python 版同一组数据):0-1:2 0-2:1 1-3:3 2-4:9 3-4:1 5-6:7
const vector<vector<int>> E7 = {{0, 1, 2}, {0, 2, 1}, {1, 3, 3}, {2, 4, 9}, {3, 4, 1}, {5, 6, 7}};
// 主连通块五个点、五条边(求最小生成树用)
const vector<vector<int>> E5 = {{0, 1, 2}, {0, 2, 1}, {1, 3, 3}, {2, 4, 9}, {3, 4, 1}};
// 有向负权反例:0→1:4 0→2:5 2→1:-3 1→3:1
const Graph NEG = {{{1, 4}, {2, 5}}, {{3, 1}}, {{1, -3}}, {}};
// 同一张图,只把 2→1 那条边改成 +3
const Graph POS = {{{1, 4}, {2, 5}}, {{3, 1}}, {{1, 3}}, {}};
// 有负环:0→1:1 1→2:-1 2→1:-1
const Graph CYC = {{{1, 1}}, {{2, -1}}, {{1, -1}}};
Graph build(int n, const vector<vector<int>>& edges) {
// 无向图:每条边 {a, b, w} 两个方向都加;邻接表按邻居编号排好
Graph g(n);
for (const auto& e : edges) {
g[e[0]].push_back({e[1], e[2]});
g[e[1]].push_back({e[0], e[2]});
}
for (auto& adj : g) sort(adj.begin(), adj.end());
return g;
}
string show(const vector<ll>& d) {
// 距离表拼成 0/2/1/...,走不到的写 -1
string s;
for (size_t i = 0; i < d.size(); i++) {
s += (i ? "/" : "") + (d[i] >= INF ? string("-1") : to_string(d[i]));
}
return s;
}
vector<ll> dijkstra(const Graph& g, int s) {
vector<ll> d(g.size(), INF);
vector<bool> done(g.size(), false);
// 小根堆:greater 让距离小的先出(不写就是大根堆)
priority_queue<pair<ll, int>, vector<pair<ll, int>>, greater<pair<ll, int>>> pq;
d[s] = 0;
pq.push({0, s});
while (!pq.empty()) {
auto [du, u] = pq.top();
pq.pop();
if (done[u]) continue;
done[u] = true;
for (auto [v, w] : g[u]) {
if (du + w < d[v]) {
d[v] = du + w;
pq.push({d[v], v});
}
}
}
return d;
}
int main() {
Graph g = build(7, E7);
cout << show(dijkstra(g, 0)) << endl;
}
(本题用 g++ -std=c++17 -O0 编译。)
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