最短路
最短路
- 1.单源最短路
- 所有边权都是正的
- 朴素Dijkstra O( n^2 )
- 堆优化Dijkstra O(mlogn)
- 存在负权边
- Bellman-Ford O(mn)
- SPFA 一般O(m),最坏O(mn)
- 所有边权都是正的
- 2.多源汇最短路 Floyd算法 O(n^3)
朴素Dijkstra
- 1.初始化距离,dis[1]=0,dis[i]=INF;
- 2.遍历集合,找到不在s(当前已确定最短距离的点)中的距离最近的点,更新当前其他点的距离;
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 510;
int m, n;
int g[N][N], 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;
}
st[t] = true;
for (int j = 1; j <= n; ++j) {
dist[j] = min(dist[j], dist[t] + g[t][j]);
}
}
if (dist[n] == 0x3f3f3f3f) return -1;
return dist[n];
}
int main() {
cin >> n >> m;
memset(g, 0x3f, sizeof g);
while (m--) {
int a, b, c;
cin >> a >> b >> c;
g[a][b] = min(g[a][b], c);
}
cout << dijkstra() << endl;
return 0;
}
堆优化Dijkstra
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
typedef pair<int, int> PII;
const int N = 100010;
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;
}
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];
}
int main () {
cin >> n >> m;
memset(h, -1, sizeof h);
while (m--) {
int a, b, c;
cin >> a >> b >> c;
add(a, b, c);
}
cout << dijkstra() << endl;
return 0;
}
Bellman-Ford
-
n次迭代,循环所有边a、b、w,更新距离;
-
dist[b] <= dist[a] + w;
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int N = 510, M = 10010;
int n, m, k;
int dist[N], backup[N];
struct {
int a, b, w;
} edges[M];
int bellman_ford() {
memset(dist, 0x3f, sizeof dist);
dist[1] = 0;
for (int i = 0; i < k; ++i) {
memcpy(backup, dist, sizeof dist);
for (int j = 0; j < m; ++j) {
int a = edges[j].a, b = edges[j].b, w = edges[j].w;
dist[b] = min(dist[b],backup[a] + w);
}
}
if (dist[n] > 0x3f3f3f3f / 2) return -1;
return dist[n];
}
int main() {
cin >> n >> m >> k;
for (int i = 0; i < m; ++i) {
int a, b, c;
cin >> a >> b >> c;
edges[i] = {a, b, c};
}
int t = bellman_ford();
if (t == -1) printf("impossible\n");
else printf("%d\n", t);
return 0;
}
SPFA
#include <iostream>
#include <algorithm>
#include <queue>
#include <cstring>
const int N = 100010;
using namespace std;
int n, m, h[N], e[N], ne[N], w[N], dist[N], idx;
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 spfa() {
memset(dist, 0x3f, sizeof dist);
dist[1] = 0;
queue<int> q;
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;
}
}
}
}
if(dist[n] == 0x3f3f3f3f) return -1;
return dist[n];
}
int main() {
cin >> n >> m;
memset(h, -1, sizeof h);
while (m--) {
int a, b, c;
cin >> a >> b >> c;
add(a, b, c);
}
int t = spfa();
if (t == -1) puts("impossible");
else cout << t << endl;
return 0;
}
Floyd
#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int N = 210, INF = 1e9;
int n, m, q;
int d[N][N];
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 >> q;
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, w;
cin >> a >> b >> w;
d[a][b] = min(d[a][b], w);
}
floyd();
while (q--) {
int a, b;
cin >> a >> b;
if (d[a][b] > INF / 2) printf("impossible\n");
else printf("%d\n", d[a][b]);
}
return 0;
}