朴素dijkstra算法

时间复杂是 O(n^2+m), n 表示点数，m 表示边数

int g[N][N];  // 存储每条边
int dist[N];  // 存储1号点到每个点的最短距离
bool st[N];   // 存储每个点的最短路是否已经确定// 求1号点到n号点的最短路，如果不存在则返回-1
int dijkstra()
{memset(dist, 0x3f, sizeof dist);dist[1] = 0;for (int i = 0; i < n - 1; i ++ ){int t = -1;     // 在还未确定最短路的点中，寻找距离最小的点for (int j = 1; j <= n; j ++ )if (!st[j] && (t == -1 || dist[t] > dist[j]))t = j;// 用t更新其他点的距离for (int j = 1; j <= n; j ++ )dist[j] = min(dist[j], dist[t] + g[t][j]);st[t] = true;}if (dist[n] == 0x3f3f3f3f) return -1;return dist[n];
}

Dijkstra求最短路 I

#include <iostream>
#include <cstring>
#include <algorithm>using namespace std;const int N = 510;int n, m;
int g[N][N];
int dist[N];
bool st[N];int dijkstra()
{memset(dist, 0x3f, sizeof dist);dist[1] = 0;for (int i = 0; i < n; i++) {int t = -1;for (int j = 1; j <= n; j++)if (!st[j] && (t == -1 || dist[t] > dist[j]))t = j;for (int j = 1; j <= n; j++)dist[j] = min(dist[j], dist[t] + g[t][j]);st[t] = true;}if (dist[n] == 0x3f3f3f3f) return  -1;return dist[n];
}int main()
{scanf("%d%d", &n, &m);memset(g, 0x3f, sizeof g);while (m--) {int a, b, c;scanf("%d%d%d", &a, &b, &c);g[a][b] = min(g[a][b], c);}printf("%d\n", dijkstra());return 0;
}

堆优化版dijkstra

时间复杂度 O(mlogn), n 表示点数，m 表示边数

typedef pair<int, int> PII;int n;      // 点的数量
int h[N], w[N], e[N], ne[N], idx;       // 邻接表存储所有边
int dist[N];        // 存储所有点到1号点的距离
bool st[N];     // 存储每个点的最短距离是否已确定// 求1号点到n号点的最短距离，如果不存在，则返回-1
int dijkstra()
{memset(dist, 0x3f, sizeof dist);dist[1] = 0;priority_queue<PII, vector<PII>, greater<PII>> heap;heap.push({0, 1});      // first存储距离，second存储节点编号while (heap.size()){auto t = heap.top();heap.pop();int ver = t.second, distance = t.first;if (st[ver]) continue;st[ver] = true;for (int i = h[ver]; i != -1; i = ne[i]){int j = e[i];if (dist[j] > distance + w[i]){dist[j] = distance + w[i];heap.push({dist[j], j});}}}if (dist[n] == 0x3f3f3f3f) return -1;return dist[n];
}

Dijkstra求最短路 II

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>using namespace std;typedef pair<int, int> PII;const int N = 150000 + 10;int n, m;
int h[N], w[N], e[N], ne[N], idx;
int dist[N];
bool st[N];void add(int a, int b, int c)
{e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
}int dijkstra()
{memset(dist, 0x3f, sizeof dist);dist[1] = 0;priority_queue<PII, vector<PII>, greater<PII> > heap;heap.push({0, 1});while (heap.size()) {auto t = heap.top();heap.pop();int ver = t.second, distance = t.first;if(st[ver]) continue;st[ver] = true;for (int i = h[ver]; i != -1; i = ne[i]) {int j = e[i];if (dist[j] > dist[ver] + w[i]) {dist[j] = dist[ver] + w[i];heap.push({dist[j], j});}}}if (dist[n] == 0x3f3f3f3f) return -1;return dist[n];
}int main()
{memset(h, -1, sizeof h);scanf("%d%d", &n, &m);while (m--) {int a, b, c;scanf("%d%d%d", &a, &b, &c);add(a, b, c);}printf("%d\n", dijkstra());return 0;
}