hdoj2544 最短路（Dijkstra || Floyd || SPFA）

题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=2544

思路

最短路算法模板题，求解使用的Dijkstra算法、Floyd算法、SPFA算法可以当做求解最短路问题的模板使用。

代码

Dijkstra算法：

 1 #include <algorithm>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <cstdio>
 5 using namespace std;
 6
 7 const int INF = 0x3f3f3f;
 8 const int N = 100 + 10;
 9 int map[N][N];
10 int dist[N];
11 int visit[N];
12 int n, m;
13
14 void dijkstra()
15 {
16     memset(visit, 0, sizeof(visit));
17     for (int i = 1; i <= n; i++)
18         dist[i] = map[1][i];
19     dist[1] = 0;
20     int min_dist, now;
21     for (int i = 1; i <= n; i++)
22     {
23         min_dist = INF;
24         for (int j = 1; j <= n; j++)
25         {
26             if (!visit[j] && dist[j] < min_dist)
27             {
28                 min_dist = dist[j];
29                 now = j;
30             }
31         }
32         if (min_dist == INF) break;
33         visit[now] = 1;
34         for (int j = 1; j <= n; j++)
35         {
36             if (dist[now] + map[now][j] < dist[j])
37                 dist[j] = dist[now] + map[now][j];
38         }
39     }
40     printf("%d\n", dist[n]);
41 }
42
43 int main()
44 {
45     //freopen("hdoj2544.txt", "r", stdin);
46     while (scanf("%d%d", &n, &m) == 2 && n)
47     {
48         memset(map, INF, sizeof(map));
49         int a, b, c;
50         for (int i = 0; i < m; i++)
51         {
52             scanf("%d%d%d", &a, &b, &c);
53             map[a][b] = map[b][a] = c;
54         }
55         dijkstra();
56     }
57     return 0;
58 }

Floyd算法：

 1 #include <algorithm>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <cstdio>
 5 using namespace std;
 6
 7 const int INF = 0x3f3f3f;
 8 const int N = 100 + 10;
 9 int map[N][N];
10 int n, m;
11
12 void floyd()
13 {
14     for (int k = 1; k <= n; k++)
15         for (int i = 1; i <= n; i++)
16             for (int j = 1;j <= n; j++)
17                 map[i][j] = min(map[i][j], map[i][k] + map[k][j]);
18     printf("%d\n", map[1][n]);
19 }
20
21 int main()
22 {
23     //freopen("hdoj2544.txt", "r", stdin);
24     while (scanf("%d%d", &n, &m) == 2 && n)
25     {
26         memset(map, INF, sizeof(map));
27         int a, b, c;
28         for (int i = 0; i < m; i++)
29         {
30             scanf("%d%d%d", &a, &b, &c);
31             map[a][b] = map[b][a] = c;
32         }
33         floyd();
34     }
35     return 0;
36 }

SPAF算法：

 1 #include <algorithm>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <cstdio>
 5 #include <queue>
 6 #include <vector>
 7 using namespace std;
 8
 9 struct Edge
10 {
11     int s, e, dist;    //边的起点、终点、长度
12
13     Edge() {}
14     Edge(int s, int e, int d) :s(s), e(e), dist(d) {}
15 };
16
17 const int INF = 0x3f3f3f;
18 const int N = 100 + 10;
19 vector<Edge> v[N];    //使用邻接表存储图
20 int dist[N];
21 int visit[N];
22 int n, m;
23
24 void spfa(int s)
25 {
26     queue<int> q;
27     memset(dist, INF, sizeof(dist));
28     memset(visit, 0, sizeof(visit));
29     q.push(s);
30     visit[s] = 1;
31     dist[s] = 0;
32
33     while (!q.empty())
34     {
35         int s = q.front();
36         q.pop();
37         visit[s] = 0;
38         for (int i = 0; i < v[s].size(); i++)
39         {
40             int e = v[s][i].e;
41             if (dist[e] > dist[s] + v[s][i].dist)
42             {
43                 dist[e] = dist[s] + v[s][i].dist;
44                 if (visit[e] == 0)
45                 {
46                     visit[e] = 1;
47                     q.push(e);
48                 }
49             }
50         }
51     }
52     printf("%d\n", dist[n]);
53 }
54
55 int main()
56 {
57     //freopen("hdoj2544.txt", "r", stdin);
58     while (scanf("%d%d", &n, &m) == 2 && n)
59     {
60         for (int i = 1;i <= n; i++)
61             v[i].clear();
62         int a, b, c;
63         for (int i = 0; i < m; i++)
64         {
65             scanf("%d%d%d", &a, &b, &c);
66             v[a].push_back(Edge(a, b, c));
67             v[b].push_back(Edge(b, a, c));
68         }
69         spfa(1);    //求结点1到其余各点的最短路径
70     }
71     return 0;
72 }