HD1599 find the mincost route(floyd + 最小环)

Posted 2020-07-04 zhaop

题目链接

题意：求最小环

第一反应时floyd判断，但是涉及到最少3个点，然后就不会了，又想的是 双联通分量，这个不知道为什么不对。

Floyd 判断 最小环

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstring>
 5 using namespace std;
 6 const int INF = 0x3f3f3f3f;
 7 const int Max = 100 + 10;
 8 int g[Max][Max], dist[Max][Max];
 9 //dist【i】【j】保存i到j的最短路经，然后i -> j -> k 就可以枚举k， dist[i][j] + g[i][k] + g[j][k]就是一个环的权值
10 int mincost;
11 void Floyed(int n)
12 {
13     mincost = INF;
14     for (int k = 1; k <= n; k++)
15     {
16         for (int i = 1; i < k; i++)
17         {
18             for (int j = i + 1; j < k; j++)
19             {
20                 if (dist[i][j] != INF && g[i][k] != INF && g[k][j] != INF)
21                 {
22                     int temp = dist[i][j] + g[i][k] + g[k][j]; // 原先这里直接相加，一直没找到错误，爆精度
23                     if (temp < mincost)
24                         mincost = temp;
25                 }
26             }
27         }
28         for (int i = 1; i <= n; i++)
29         {
30             for (int j = 1; j <= n; j++)
31             {
32                 if (dist[i][k] != INF && dist[k][j] != INF)
33                     dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
34             }
35         }
36     }
37 }
38 int main()
39 {
40     int n, m;
41     while (scanf("%d%d", &n, &m) != EOF)
42     {
43         for (int i = 1; i < Max; i++)
44             for (int j = 1; j < Max; j++)
45             {
46                 g[i][j] = INF;
47                 dist[i][j] = INF;
48             }
49         int a, b, c;
50         for (int i = 1; i <= m; i++)
51         {
52             scanf("%d%d%d", &a, &b, &c);
53             if (g[a][b] > c)
54                 g[a][b] = g[b][a] = dist[a][b] = dist[b][a] = c;
55         }
56         Floyed(n);
57         if (mincost >= INF)
58             printf("It\'s impossible.\\n");
59         else
60             printf("%d\\n", mincost);
61 
62     }
63     return 0;
64 }

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