POJ 2253 Frogger

时间：2014-11-01 11:35:32 阅读：250 评论：0 收藏：0 [点我收藏+]

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Frogger

Time Limit: 1000ms

Memory Limit: 65536KB

This problem will be judged on PKU. Original ID: 2253
64-bit integer IO format: %lld Java class name: Main

Freddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on another stone. He plans to visit her, but since the water is dirty and full of tourists‘ sunscreen, he wants to avoid swimming and instead reach her by jumping.
Unfortunately Fiona‘s stone is out of his jump range. Therefore Freddy considers to use other stones as intermediate stops and reach her by a sequence of several small jumps.
To execute a given sequence of jumps, a frog‘s jump range obviously must be at least as long as the longest jump occuring in the sequence.
The frog distance (humans also call it minimax distance) between two stones therefore is defined as the minimum necessary jump range over all possible paths between the two stones.

You are given the coordinates of Freddy‘s stone, Fiona‘s stone and all other stones in the lake. Your job is to compute the frog distance between Freddy‘s and Fiona‘s stone.

Input

The input will contain one or more test cases. The first line of each test case will contain the number of stones n (2<=n<=200). The next n lines each contain two integers xi,yi (0 <= xi,yi <= 1000) representing the coordinates of stone #i. Stone #1 is Freddy‘s stone, stone #2 is Fiona‘s stone, the other n-2 stones are unoccupied. There‘s a blank line following each test case. Input is terminated by a value of zero (0) for n.

Output

For each test case, print a line saying "Scenario #x" and a line saying "Frog Distance = y" where x is replaced by the test case number (they are numbered from 1) and y is replaced by the appropriate real number, printed to three decimals. Put a blank line after each test case, even after the last one.

Sample Input

Sample Output

Scenario #1
Frog Distance = 5.000

Scenario #2
Frog Distance = 1.414

Source

Ulm Local 1997

解题：最小生成树算法以及 dijkstra Floyd都可以啊。

先上Prim算法版

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <climits>
 7 #include <vector>
 8 #include <queue>
 9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define pii pair<int,int>
15 #define INF 0x3f3f3f3f
16 using namespace std;
17 const int maxn = 210;
18 double dis[maxn][maxn],d[maxn];
19 bool vis[maxn];
20 vector<int>g[maxn];
21 int n,x[maxn],y[maxn],p[maxn];
22 void prim() {
23     for(int i = 0; i <= n; ++i) {
24         vis[i] = false;
25         d[i] = INF;
26         p[i] = -1;
27     }
28     d[1] = 0;
29     while(true) {
30         double minV = INF;
31         int index = -1;
32         for(int i = 1; i <= n; ++i)
33             if(!vis[i] && d[i] < minV) minV = d[index = i];
34         if(index == -1 || minV >= INF) break;
35         if(p[index] > -1) {
36             g[p[index]].push_back(index);
37             g[index].push_back(p[index]);
38         }
39         vis[index] = true;
40         for(int i = 1; i <= n; ++i)
41             if(!vis[i] && d[i] > dis[index][i]) {
42                 d[i] = dis[index][i];
43                 p[i] = index;
44             }
45     }
46 }
47 double getDis(int i,int j) {
48     double tmp = (x[i] - x[j])*(x[i] - x[j]);
49     tmp += (y[i] - y[j])*(y[i] - y[j]);
50     return sqrt(tmp);
51 }
52 double ans;
53 bool dfs(int u,int fa,double maxV) {
54     if(u == 2) {
55         ans = maxV;
56         return true;
57     }
58     for(int i = g[u].size()-1; i >= 0; --i) {
59         if(g[u][i] == fa) continue;
60         if(dfs(g[u][i],u,max(maxV,dis[u][g[u][i]]))) return true;
61     }
62     return false;
63 }
64 int main() {
65     int cs = 1;
66     while(scanf("%d",&n),n) {
67         for(int i = 0; i <= n; ++i) g[i].clear();
68         for(int i = 1; i <= n; ++i)
69             scanf("%d %d",x+i,y+i);
70         for(int i = 1; i <= n; ++i)
71             for(int j = 1; j <= n; ++j)
72                 dis[i][j] = getDis(i,j);
73         prim();
74         ans = 0;
75         dfs(1,-1,0);
76         printf("Scenario #%d\nFrog Distance = %.3f\n\n",cs++,ans);
77     }
78     return 0;
79 }

View Code

Dijkstra版

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <climits>
 7 #include <vector>
 8 #include <queue>
 9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define pii pair<int,int>
15 #define INF 0x3f3f3f3f
16 using namespace std;
17 const int maxn = 210;
18 double dis[maxn][maxn],d[maxn];
19 int n,x[maxn],y[maxn];
20 bool vis[maxn];
21 double getDis(int i,int j) {
22     double tmp = (x[i] - x[j])*(x[i] - x[j]);
23     tmp += (y[i] - y[j])*(y[i] - y[j]);
24     return sqrt(tmp);
25 }
26 void dijkstra(){
27     for(int i = 1; i <= n; ++i){
28         vis[i] = false;
29         d[i] = INF;
30     }
31     d[1] = 0;
32     while(true){
33         double minV = INF;
34         int index = -1;
35         for(int i = 1; i <= n; ++i)
36             if(!vis[i] && d[i] < minV) minV = d[index = i];
37         if(index == -1 || minV >= INF) break;
38         vis[index] = true;
39         for(int i = 1; i <= n; ++i)
40             if(!vis[i] && d[i] > max(d[index],dis[index][i]))
41                 d[i] = max(d[index],dis[index][i]);
42     }
43 }
44 int main() {
45     int cs = 1;
46     while(scanf("%d",&n),n) {
47         for(int i = 1; i <= n; ++i)
48             scanf("%d %d",x+i,y+i);
49         for(int i = 1; i <= n; ++i)
50             for(int j = 1; j <= n; ++j)
51                 dis[i][j] = getDis(i,j);
52         dijkstra();
53         printf("Scenario #%d\nFrog Distance = %.3f\n\n",cs++,d[2]);
54     }
55     return 0;
56 }

View Code

Floyd版

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <climits>
 7 #include <vector>
 8 #include <queue>
 9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define pii pair<int,int>
15 #define INF 0x3f3f3f3f
16 using namespace std;
17 const int maxn = 210;
18 double dis[maxn][maxn];
19 int n,x[maxn],y[maxn];
20 double getDis(int i,int j) {
21     if(i == j) return 0;
22     double tmp = (x[i] - x[j])*(x[i] - x[j]);
23     tmp += (y[i] - y[j])*(y[i] - y[j]);
24     return sqrt(tmp);
25 }
26 void Floyd(){
27     for(int k = 1; k <= n; ++k){
28         for(int i = 1; i <= n; ++i){
29             for(int j = 1; j <= n; ++j)
30                 dis[i][j] = min(dis[i][j],max(dis[i][k],dis[k][j]));
31         }
32     }
33 }
34 int main() {
35     int cs = 1;
36     while(scanf("%d",&n),n) {
37         for(int i = 1; i <= n; ++i)
38             scanf("%d %d",x+i,y+i);
39         for(int i = 1; i <= n; ++i)
40             for(int j = 1; j <= n; ++j)
41                 dis[i][j] = getDis(i,j);
42         Floyd();
43         printf("Scenario #%d\nFrog Distance = %.3f\n\n",cs++,dis[1][2]);
44     }
45     return 0;
46 }

View Code

POJ 2253 Frogger

标签：style blog http io color os ar java for

原文地址：http://www.cnblogs.com/crackpotisback/p/4066471.html

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