题解 | #【模板】单源最短路Ⅲ ‖ 非负权图#
【模板】单源最短路Ⅲ ‖ 非负权图
https://www.nowcoder.com/practice/d7fafd4f3340439e90597532850257b5
一样的代码,改一下大小和输入权值就行
#include <iostream>
#include <queue>
#include <map>
#include <set>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <numeric>
#include <ctime>
#include <string>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
using namespace std;
using ll = long long;
const ll N = 5e5 + 5, mod = 1e9 + 7, inf = 2e18;
const double esp = 1e-8;
int n, m, s;
struct Node {
ll v, w;
};
vector<Node>g[N];
bool vis[N];
ll dis[N];
struct node {
ll id, vlu;
bool operator<(const node& u)const {
return u.vlu < vlu;
}
};
void dijk(int s) {
for (int i = 0; i <= n + 3; i++) {
dis[i] = inf;
}
priority_queue<node>pq;
pq.push({s, dis[s] = 0});
while (!pq.empty()) {
node tem = pq.top();
pq.pop();
if (vis[tem.id])continue;
vis[tem.id] = true;
for (auto [y, w] : g[tem.id]) {
if (dis[y] > dis[tem.id] + w) {
pq.push({y, dis[y] = dis[tem.id] + w});
}
}
}
}
void solve() {
cin >> n >> m >> s;
while (m--) {
ll u, v, w;
cin >> u >> v >> w;
g[u].push_back({v, w});
}
dijk(s);
for (int i = 1; i <= n; i++) {
cout << (dis[i] == inf ? -1 : dis[i]) << " \n"[i == n];
}
}
int main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int t = 1;
//cin >> t;
while (t--) {
solve();
}
return 0;
}


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