POJ 2253 Frogger（最短路变形）

题目链接

这个题就是找到1 到 2之间的路径上所有经过路的最大值中的最小值。。。有点绕，反正就是最短路的变形。。。没太想清楚怎么写，写了spfa和floyd都错了，没想清楚怎么变形啊。。。

一个地方想错了。。。改了都过了。。

 1 #include <cstdio>//spfa算法

 2 #include <cstring>

 3 #include <cmath>

 4 #include <queue>

 5 using namespace std;

 6 #define INF 0x3f3f3f3f

 7 #define eps 1e-7

 8 double dp[201][201],d[201];

 9 int x[201],y[201],n;

10 double judge(int i,int j)

11 {

12     return sqrt((x[i]-x[j])*(x[i]-x[j])*1.0+(y[i]-y[j])*(y[i]-y[j])*1.0);

13 }

14 void spfa()

15 {

16     int u,i;

17     int in[201];

18     queue<int> que;

19     memset(in,0,sizeof(in));

20     for(i = 1;i <= n;i ++)

21     {

22         d[i] = INF;

23     }

24     d[1] = 0;

25     que.push(1);

26     in[1] = 1;

27     while(!que.empty())

28     {

29         u = que.front();

30         que.pop();

31         in[u] = 0;

32         for(i = 1;i <= n;i ++)

33         {

34             if(d[i] > max(d[u],dp[u][i]))

35             {

36                 d[i] = max(d[u],dp[u][i]);

37                 if(!in[i])

38                 {

39                     in[i] = 1;

40                     que.push(i);

41                 }

42             }

43         }

44     }

45 }

46 int main()

47 {

48     int i,j,cas = 0;

49     while(scanf("%d",&n)!=EOF)

50     {

51         cas ++;

52         if(n == 0) break;

53         for(i = 1;i <= n;i ++)

54         scanf("%d%d",&x[i],&y[i]);

55         for(i = 1;i <= n;i ++)

56         {

57             for(j = 1;j <= n;j ++)

58             dp[i][j] = judge(i,j);

59         }

60         spfa();

61         printf("Scenario #%d\nFrog Distance = ",cas);

62         printf("%.3lf\n\n",d[2]);

63     }

64     return 0;

65 }

 1 #include <cstdio>//floyd算法

 2 #include <cstring>

 3 #include <cmath>

 4 #include <queue>

 5 using namespace std;

 6 #define INF 0x3f3f3f3f

 7 #define eps 1e-7

 8 double dp[201][201];

 9 int x[201],y[201],n;

10 double judge(int i,int j)

11 {

12     return sqrt((x[i]-x[j])*(x[i]-x[j])*1.0+(y[i]-y[j])*(y[i]-y[j])*1.0);

13 }

14 int main()

15 {

16     int i,j,k,cas = 0;

17     while(scanf("%d",&n)!=EOF)

18     {

19         cas ++;

20         if(n == 0) break;

21         for(i = 1;i <= n;i ++)

22         scanf("%d%d",&x[i],&y[i]);

23         for(i = 1;i <= n;i ++)

24         {

25             for(j = 1;j <= n;j ++)

26             dp[i][j] = judge(i,j);

27         }

28         for(i = 1;i <= n;i ++)

29         {

30             for(j = 1;j <= n;j ++)

31             {

32                 for(k = 1;k <= n;k ++)

33                 {

34                     if(dp[j][k] > max(dp[j][i],dp[i][k])&&i != j&&j != k&&i != k)

35                     {

36                         dp[j][k] = max(dp[j][i],dp[i][k]);

37                     }

38                 }

39             }

40         }

41         printf("Scenario #%d\nFrog Distance = ",cas);

42         printf("%.3lf\n\n",dp[1][2]);

43     }

44     return 0;

45 }