背景
- 今天基本上刷了所有的最短路径算法了,从之前无权重的BFS,然后是图论里的最短路径算法。acwing上总结了5种算法的用法,如下图:
-
朴素dijkstra算法O(n ^ 2),适合稠密图
朴素dijkstra算法的思路是,初始化距离d[1] = 0, 其余为正无穷INF。
- 循环n - 次,从当前距离数组中选出距离最短且没有遍历过的节点i,并用该节点的出边w更新其余的邻接点j的距离,d[j] = min(d[j], d[i] + w)
- 最后d[n]就是1号节点到所有节点的最短路径。
```cpp
include
include
using namespace std;
define N 510
int g[N][N]; int dist[N]; bool st[N]; int n, m;
void 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]) {
continue;
}
if (t == -1 || dist[j] < dist[t]) {
t = j;
}
}
// cout << t << " " << dist[t] << endl;
st[t] = true;
for (int j = 1; j <= n; j ++) {
dist[j] = min(dist[j], dist[t] + g[t][j]);
}
}
}
int main() { scanf(“%d%d”, &n, &m); memset(g, 0x3f, sizeof g);
while(m --) {
int a, b, d;
scanf("%d%d%d", &a, &b, &d);
g[a][b] = min(g[a][b], d);
}
dijkstra();
if (dist[n] == 0x3f3f3f3f) {
cout << -1 << endl;
} else {
cout << dist[n] << endl;
}
return 0;
}
<a name="J3R57"></a>
#### 堆优化的dijkstra算法O(m logn),适合稀疏图
- 相比朴素dijkstra算法,堆优化的算法是通过堆存放距离,并且可以logm的复杂度获取对应的最短距离。
- 其余保持不变
```cpp
#include <iostream>
#include <cstring>
#include <queue>
#include <vector>
using namespace std;
#define M 150010
int dist[M], h[M];
bool st[M];
int e[M], ne[M], w[M], idx;
int n, m;
typedef pair<int, int> PII;
void add(int a, int b, int d)
{
e[idx] = b;
ne[idx] = h[a];
w[idx] = d;
h[a] = idx;
idx ++;
}
int main()
{
scanf("%d%d", &n, &m);
memset(h, -1, sizeof(h));
memset(dist, 0x3f, sizeof(dist));
dist[1] = 0;
while(m --) {
int a, b, d;
scanf("%d%d%d", &a, &b, &d);
add(a, b, d);
}
priority_queue<PII, vector<PII>, greater<PII>> heap;
// first represent distance
// second represent node idx
heap.push({0, 1});
while(heap.size()) {
auto t = heap.top();
heap.pop();
if (st[t.second]) {
continue;
}
st[t.second] = true;
for (int i = h[t.second]; i != -1; i = ne[i]) {
if (dist[e[i]] > dist[t.second] + w[i]) {
dist[e[i]] = dist[t.second] + w[i];
heap.push({dist[e[i]], e[i]});
}
}
}
if (dist[n] == 0x3f3f3f3f) {
cout << -1 << endl;
} else {
cout << dist[n] << endl;
}
return 0;
}
bellman-ford算法O(nm)
- bellman-ford算法的思路是,循环n次,每次更新都遍历所有的边,看哪些距离可以更新,能更新的就更新下。循环n次后,d[n]就是1号节点到所有节点的最短路径。
- bellman-ford算法的循环次数的实际含义是,经过k条边的最短路径。
- 根据循环次数的定义,该次循环只需要更新加一条边后对应的距离,backup数组是为了防止连续更新。
```cpp
include
include
using namespace std;
const int N = 510, M = 10010; struct Edge { int a, b, w; } edge[M]; int n, m, k; int dist[N], back_dist[N];
void bellman_ford() { memset(dist, 0x3f, sizeof(dist)); dist[1] = 0;
for (int i = 0; i < k; i ++) {
memcpy(back_dist, dist, sizeof(dist));
for (int j = 0; j < m; j ++) {
int a = edge[j].a, b = edge[j].b, w = edge[j].w;
dist[b] = min(dist[b], back_dist[a] + w);
}
}
}
int main() {
scanf("%d%d%d", &n, &m, &k);
for (int i = 0; i < m; i ++) {
scanf("%d%d%d", &edge[i].a, &edge[i].b, &edge[i].w);
}
bellman_ford();
if (dist[n] > 0x3f3f3f3f / 2) {
puts("impossible");
} else {
cout << dist[n] << endl;
}
return 0;
}
<a name="KDvku"></a>
#### spfa算法O(m),最坏O(nm)
- 其实根据dijkstra算法可以知道,更新距离的时候并不需要查看所有的边。
- 更新边的必要条件是前一个节点的的距离更新了,所以可以维护一个节点距离更新的队列,每个节点距离更新后就放在队列中,然后每次拿一个节点来更新邻接点的距离。
- 队列空的时候d[n]就是1号节点到所有节点的最短路径。
- st数组标识是否在队列中,防止重复更新
```cpp
#include <iostream>
#include <cstring>
#include <queue>
using namespace std;
#define N 100010
int h[N], e[N], ne[N], w[N], dist[N];
bool st[N];
int idx, n, m;
void add(int a, int b, int c)
{
e[idx] = b;
ne[idx] = h[a];
h[a] = idx;
w[idx] = c;
idx ++;
}
void spfa()
{
memset(dist, 0x3f, sizeof dist);
queue<int> q;
dist[1] = 0;
q.push(1);
st[1] = true;
while(q.size()) {
int t = q.front();
q.pop();
st[t] = false;
for (int i = h[t]; i != -1; i = ne[i]) {
int j = e[i];
if (dist[j] > dist[t] + w[i]) {
dist[j] = dist[t] + w[i];
if (!st[j]) {
q.push(j);
st[j] = true;
}
}
}
}
}
int main()
{
memset(h, -1, sizeof h);
scanf("%d%d", &n, &m);
for (int i = 0; i < m; i ++) {
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
add(a, b, c);
}
spfa();
if (dist[n] == 0x3f3f3f3f) {
cout << "impossible" << endl;
} else {
cout << dist[n] << endl;
}
return 0;
}
Floyd算法
const int N = 210, M = 20010, 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() { scanf(“%d%d%d”, &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;
scanf("%d%d%d", &a, &b, &c);
d[a][b] = min(d[a][b], c);
}
floyd();
while(k --) {
int a, b;
scanf("%d%d", &a, &b);
if (d[a][b] > INF / 2) {
cout << "impossible" << endl;
} else {
cout << d[a][b] << endl;
}
}
return 0;
} ```