HDU 2066 一个人的旅行

最短路算法优化与应用

最新推荐文章于 2023-03-19 16:29:49 发布

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本文探讨了使用优先队列优化的Dijkstra算法解决从多个起点到多个终点的最短路径问题，并提供了两种实现方式：一种使用二维数组，另一种使用邻接表。通过实例演示了如何将超级源点和超级汇点引入算法中，简化路径查找过程。最后，通过代码实现了0ms的优化版本，展示了高效求解策略。

最短路，求从k1个初始点到 k2个终点中最短的一条路径长度。

设置一个超级源点和超级汇点即可，使得超级源点到k1个初始点的权值为0，超级汇点. . .

下面是用priority_queue优化的Dijkstra代码：

78ms。

#include <iostream> #include <cstring> #include <cstdio> #include <queue> #include <algorithm> using namespace std; #define MAXN 1010 #define INF 0x3fffffff int map[MAXN][MAXN]; int dis[MAXN]; bool hash[MAXN]; struct Dij { int id; int dis; bool friend operator < (const Dij a, const Dij b) { return a.dis > b.dis; } }; Dij now, next; void Init() { for (int i = 0; i < MAXN; i++) { map[i][i] = 0; dis[i] = INF; for (int j = i + 1; j < MAXN; j++) { map[i][j] = map[j][i] = INF; } } } int Bfs(int n) { priority_queue<Dij> Q; now.id = 0; now.dis = 0; dis[0] = 0; Q.push(now); memset(hash, false, sizeof(hash)); while (!Q.empty()) { now = Q.top(); Q.pop(); if (hash[now.id]) { continue; } if (now.id == n) { return now.dis; } hash[now.id] = true; for (int j = 0; j <= n; j++) { if (!hash[j] && dis[now.id] + map[now.id][j] < dis[j]) { dis[j] = dis[now.id] + map[now.id][j]; next.id = j; next.dis = dis[j]; Q.push(next); } } } } int main() { int t, s, d, a, b, c, max, ans, i, ss, dd; while (scanf("%d %d %d", &t, &s, &d) != EOF) { Init(); int max = 0; for (i = 0; i < t; i++) { scanf("%d %d %d", &a, &b, &c); if (c < map[a][b]) { map[a][b] = map[b][a] = c; } if (a > max) max = a; if (b > max) max = b; } while (s--) { scanf("%d", &ss); map[0][ss] = map[ss][0] = 0; } while (d--) { scanf("%d", &dd); map[dd][max + 1] = map[max + 1][dd] = 0; } ans = Bfs(max + 1); printf("%d/n", ans); } return 0; }

下面用邻接表写的 0ms。

#include <iostream> #include <cstring> #include <cstdio> #include <queue> #include <algorithm> using namespace std; #define MAXN 1010 #define INF 0x3fffffff bool hash[MAXN]; struct Edge {//边 int adj; int weight; Edge *next; }; Edge *map[MAXN], vex[MAXN];//vex顶点 struct Dij { int id; int dis; bool friend operator < (const Dij a, const Dij b) { return a.dis > b.dis; } }; Dij now, next; void Init() { for (int i = 0; i < MAXN; i++) { map[i] = &vex[i];//初始指向顶点 } } void Creat(int a, int b, int c) {//建邻接表与a邻接是b Edge *pa = new Edge; pa->adj = b; pa->weight = c; pa->next = NULL; map[a]->next = pa; map[a] = map[a]->next; } int Bfs(int st, int ed, int n) { memset(hash, false, sizeof(hash)); priority_queue<Dij> Q; now.id = st; now.dis = st; Q.push(now); while (!Q.empty()) { now = Q.top(); Q.pop(); if (hash[now.id]) { continue; } if (now.id == ed) { return now.dis; } hash[now.id] = true; Edge *first = vex[now.id].next; for (Edge *j = first; j != NULL; j = j->next) { if (hash[j->adj]) { continue; } next.id = j->adj; next.dis = now.dis + j->weight; Q.push(next); } } } int main() { int i, n, t, s, d, ss, dd, a, b, c; while (scanf("%d %d %d", &t, &s, &d) != EOF) { Init(); n = 0; for (i = 0; i < t; i++) { scanf("%d %d %d", &a, &b, &c); Creat(a, b, c); Creat(b, a, c); if (a > n) { n = a; } if (b > n) { n = b; } } while (s--) { scanf("%d", &ss); Creat(0, ss, 0); Creat(ss, 0, 0); } while (d--) { scanf("%d", &dd); Creat(dd, n + 1, 0); Creat(n + 1, dd, 0); } int ans = Bfs(0, n + 1, n + 1); printf("%d/n", ans); } return 0; }