Acwing-----演算法基礎課之第三章搜尋與圖論(二)
阿新 • • 發佈:2020-09-20
最短路:
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1.單源最短路
- 所有邊權都是正數:樸素Dijkstra演算法(O(\(n^2\)))、堆優化Dijkstra演算法(O(\(mlongn\)))
- 存在負權邊:Bellman-Ford演算法(O(\(nm\)))、SPFA演算法(一般O(\(m\)),最壞O(\(nm\)))
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2.多源匯最短路:Floyd演算法(O(\(n^3\)))
849. Dijkstra求最短路 I
// 樸素版Dijkstra #include <iostream> #include <algorithm> #include <cstring> using namespace std; const int N = 510; int n, m, 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; }
850. Dijkstra求最短路 II
// 堆優化 #include <iostream> #include <algorithm> #include <cstring> #include <queue> using namespace std; typedef pair<int, int> PII; const int N = 1e6 + 10; int n, m, h[N], e[N], ne[N], dist[N], w[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 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, d = 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] > d + w[i]) { dist[j] = d + 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; }
853. 有邊數限制的最短路
#include <iostream> #include <algorithm> #include <cstring> using namespace std; const int N = 510, M = 100010; int n, m, k, dist[N], backup[N]; struct edge { int a, b, c; } edges[M]; void 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) { auto e = edges[j]; dist[e.b] = min(dist[e.b], backup[e.a] + e.c); } } } 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}; } bellman_ford(); if (dist[n] > 0x3f3f3f3f / 2) puts("impossible"); else cout << dist[n] << endl; return 0; }
851. spfa求最短路
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
using namespace std;
const int N = 100010;
int n, m, e[N], ne[N], h[N], w[N], idx, 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++;
}
void 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;
}
}
}
}
}
int main() {
cin >> n >> m;
memset(h, -1, sizeof h);
while (m--) {
int a, b, c;
cin >> a >> b >> c;
add(a, b, c);
}
spfa();
if (dist[n] == 0x3f3f3f3f) puts("impossible");
else cout << dist[n] << endl;
return 0;
}
852. spfa判斷負環
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int N = 2010, M = 10010;
int n, m, e[M], h[N], w[M], dist[N], ne[M], idx, 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) {
st[i] = true;
q.push(i);
}
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];
cnt[j] = cnt[t] + 1;
if (cnt[j] >= n) return true;
if (!st[j]) {
q.push(j);
st[j] = true;
}
}
}
}
return false;
}
int main() {
cin >> n >> m;
memset(h, -1, sizeof h);
while (m--) {
int a, b, c;
cin >> a >> b >> c;
add(a, b, c);
}
if (spfa()) puts("Yes");
else puts("No");
return 0;
}
854. Floyd求最短路
#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, c;
cin >> a >> b >> c;
d[a][b] = min(d[a][b], c);
}
floyd();
while (q--) {
int a, b;
cin >> a >> b;
if (d[a][b] > INF / 2) puts("impossible");
else cout << d[a][b] << endl;
}
return 0;
}