POJ-2253 Frogger---最短路变形&&最大边的最小值

Posted 2020-10-29 努力努力再努力x

题目链接：

https://vjudge.net/problem/POJ-2253

题目大意：

青蛙A想访问青蛙B，必须跳着石头过去，不幸的是，B所在的石头太远了，需要借助其他的石头，求从A到B的路径中，青蛙最少需要的跳跃能力是多远

思路：

理清题意，这里规定的是每条路中的最大边为青蛙需要的跳跃能力，要求这个跳跃能力的最小值，思路和POJ2263一样POJ2263求的是最小边的最大值，这里求的是最大边的最小值，同样是松弛方程改变一下就可以，三种算法均可，这里附上Floyd和dijkstra

Floyd：188ms

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<cmath>
 6 #include<queue>
 7 #include<stack>
 8 #include<map>
 9 #include<set>
10 #include<sstream>
11 #define MEM(a, b) memset(a, b, sizeof(a));
12 using namespace std;
13 typedef long long ll;
14 const int maxn = 200 + 10;
15 const int INF = 0x3f3f3f3f;
16 int T, n, m, cases, tot;
17 double Map[maxn][maxn];
18 struct node
19 {
20     double x, y;
21 };
22 node a[maxn];
23 int main()
24 {
25     while(cin >> n && n)
26     {
27         MEM(a, 0);
28         MEM(Map, 0);
29         for(int i = 1; i <= n; i++)
30         {
31             cin >> a[i].x >> a[i].y;
32         }
33         for(int i = 1; i <= n; i++)
34         {
35             for(int j = 1; j <= n; j++)
36                 Map[i][j] = sqrt((a[i].x - a[j].x) * (a[i].x - a[j].x) + (a[i].y - a[j].y) * (a[i].y - a[j].y));
37         }
38         for(int k = 1; k <= n; k++)
39         {
40             for(int i = 1; i <= n; i++)
41             {
42                 for(int j = 1; j <= n; j++)
43                 {
44                     Map[i][j] = min(Map[i][j], max(Map[i][k], Map[k][j]));
45                 }
46             }
47         }
48         printf("Scenario #%d\\n", ++cases);
49         printf("Frog Distance = %.3f\\n\\n", Map[1][2]);
50     }
51     return 0;
52 }

dijkstra：47ms

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<cmath>
 6 #include<queue>
 7 #include<stack>
 8 #include<map>
 9 #include<set>
10 #include<sstream>
11 #define MEM(a, b) memset(a, b, sizeof(a));
12 using namespace std;
13 typedef long long ll;
14 const int maxn = 200 + 10;
15 const int INF = 0x3f3f3f3f;
16 int T, n, m, cases, tot;
17 double Map[maxn][maxn];
18 struct node
19 {
20     double x, y;
21 };
22 node a[maxn];
23 bool v[maxn];
24 double d[maxn];
25 void dijkstra(int u)
26 {
27     memset(v, 0, sizeof(v));
28     for(int i = 0; i <= n; i++)d[i] = INF;
29     d[u] = 0;
30     for(int i = 1; i <= n; i++)
31     {
32         int x, m = INF;
33         for(int i = 1; i <= n; i++)if(!v[i] && d[i] <= m)m = d[x = i];//找最小值
34         v[x] = 1;
35         for(int i = 1; i <= n; i++)d[i] = min(d[i], max(d[x], Map[x][i]));
36     }
37 }
38 int main()
39 {
40     while(cin >> n && n)
41     {
42         MEM(a, 0);
43         MEM(Map, 0);
44         for(int i = 1; i <= n; i++)
45         {
46             cin >> a[i].x >> a[i].y;
47         }
48         for(int i = 1; i <= n; i++)
49         {
50             for(int j = 1; j <= n; j++)
51                 Map[i][j] = sqrt((a[i].x - a[j].x) * (a[i].x - a[j].x) + (a[i].y - a[j].y) * (a[i].y - a[j].y));
52         }
53         dijkstra(1);
54         printf("Scenario #%d\\n", ++cases);
55         printf("Frog Distance = %.3f\\n\\n", d[2]);
56     }
57     return 0;
58 }