网站程序如何制作吉安seo招聘

图的存储方式

2、最短路问题

最短路问题可以分为单源最短路问题和多源最短路问题，单源最短路问题就是求出从点1->n的最短距离，而多源最短路问题就是求出从点i->j的最短距离。单源最短路问题还可以分为正权边的单源最短路问题和负权边的单源最短路问题。具体算法和时间复杂度如下图：

2.1、Dijkstra算法（朴素版）

算法模板：

#include <iostream>
#include <cstring>using namespace std;
const int N = 510;
int g[N][N], d[N];
int n, m;
bool st[N];int dijkstra()
{memset(d, 0x3f, sizeof d);d[1] = 0;for (int i = 0; i < n; i++){int t = -1;for (int j = 1; j <= n; j++)if (!st[j] && (t == -1 || d[t] > d[j]))t = j;st[t] = true;for (int j = 1; j <= n; j++)d[j] = min(d[j], d[t] + g[t][j]);}return d[n] == 0x3f3f3f3f ? -1 : d[n];
}int main()
{cin >> 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);}cout << dijkstra() << endl;return 0;
}

2.2、Dijkstra算法（堆优化版）

下面来看看如何优化：

算法模板：

#include <iostream>
#include <cstring>
#include <queue>using namespace std;
typedef pair<int, int> PII;
const int N = 1.5e5+10;
int h[N], e[N], w[N], ne[N], idx;
int n, m, d[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(d, 0x3f, sizeof d);d[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, dis = t.first;if (st[ver]) continue;st[ver] = true;for (int i = h[ver]; i != -1; i = ne[i]){int j = e[i];if (d[j] > dis + w[i]){d[j] = dis + w[i];heap.push({d[j], j});}}}return d[n] == 0x3f3f3f3f ? -1 : d[n];
}int main()
{cin >> n >> m;memset(h, -1, sizeof h);while (m--){int a, b, c;scanf("%d%d%d", &a, &b, &c);add(a, b, c);}cout << dijkstra() << endl;return 0;
}

2.3、Bellman-Ford算法

代码模板：

#include <iostream>
#include <cstring>using namespace std;
const int N = 510, M = 10010;
int n, m, k;
int d[N], backup[N];struct Edge
{int a, b, w;
}edges[M];void bellman_ford()
{memset(d, 0x3f, sizeof d);d[1] = 0;for (int i = 0; i < k; i++){memcpy(backup, d, sizeof d);for (int j = 0; j < m; j++){auto e = edges[j];d[e.b] = min(d[e.b], backup[e.a] + e.w);}}
}int main()
{cin >> n >> m >> k;for (int i = 0; i < m; i++){int a, b, w;scanf("%d%d%d", &a, &b, &w);edges[i] = {a, b, w};}bellman_ford();if (d[n] > 0x3f3f3f3f / 2) cout << "impossible" << endl;else cout << d[n] << endl;return 0;
}

2.4、SPFA求最短路

代码模板：

#include <iostream>
#include <cstring>
#include <queue>using namespace std;
const int N = 1e5+10;
int n, m;
int h[N], e[N], w[N], ne[N], idx;
int d[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++;
}void spfa()
{memset(d, 0x3f, sizeof d);d[1] = 0;queue<int> q;q.push(1);st[1] = true;while (q.size()){auto t = q.front();q.pop();st[t] = false;for (int i = h[t]; i != -1; i = ne[i]){int j = e[i];if (d[j] > d[t] + w[i]){d[j] = d[t] + w[i];if (!st[j]){st[j] = true;q.push(j);}}}}
}int main()
{cin >> n >> m;memset(h, -1, sizeof h);while (m--){int a, b, c;scanf("%d%d%d", &a, &b, &c);add(a, b, c);}spfa();if (d[n] == 0x3f3f3f3f) cout << "impossible" << endl;else cout << d[n] << endl;return 0;
}

2.5、SPFA判负环

#include <iostream>
#include <cstring>
#include <queue>using namespace std;
const int N = 2010, M = 10010;
int h[N], e[M], w[M], ne[M], idx;
int n, m;
int d[N], cnt[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++;
}bool spfa()
{queue<int> q;for (int i = 1; i <= n; i++){q.push(i);st[i] = true;}while (q.size()){auto t = q.front();q.pop();st[t] = false;for (int i = h[t]; i != -1; i = ne[i]){int j = e[i];if (d[j] > d[t] + w[i]){d[j] = d[t] + w[i];cnt[j] = cnt[t] + 1;if (cnt[j] >= n) return true;if (!st[j]){st[j] = true;q.push(j);}}}}return false;
}int main()
{cin >> n >> m;memset(h, -1, sizeof h);while (m--){int a, b, c;scanf("%d%d%d", &a, &b, &c);add(a, b, c);}if (spfa()) cout << "Yes" << endl;else cout << "No" << endl;return 0;
}

2.6、Floyd算法

#include <iostream>
#include <cstring>using namespace std;
const int N = 210, INF = 0x3f3f3f3f;
int d[N][N];
int n, m, k;void floyd()
{for (int k = 1; k <= n; k++)for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
}int main()
{cin >> n >> m >> k;for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)if (i == j) d[i][j] = 0;else d[i][j] = INF;while (m--){int a, b, c;cin >> a >> b >> c;d[a][b] = min(d[a][b], c);}floyd();while (k--){int l, r;cin >> l >> r;if (d[l][r] > INF / 2) cout << "impossible" << endl;else cout << d[l][r] << endl;}return 0;
}