poj2253佛洛依德

题意：给出青蛙A,B和若干石头的坐标，现青蛙A想到青蛙B那，A可通过任意石头到达B,

问从A到B多条路径中的最长边中最短的边

因为n<=200,故用佛洛依德水一哈。

佛洛依德是来求多源最短路，把ai和aj两点的最短距离通过松弛来求，用另一个点k作为中间节点

i-k,k-j来松弛ij之间的距离，整体上是dp的思想.

核心代码是三个循环

本题比较特殊是求最长边里面的最短边所以要修改最后的转台转移的条件和状态转移方程

if(d[i][j]>max(d[i][k],d[k][j])d[i][j]=max(d[i][k],d[k][j]);

code:

using namespace std;
typedef long long ll;
const double INF = 0x7fffffff;
double d[215][215];
int n;
double counter(int i, int j)
{
	return sqrt((double)(corr[i].x - corr[j].x)*(corr[i].x - corr[j].x) + (corr[i].y - corr[j].y)*(corr[i].y - corr[j].y));
}
int main()
{
	int iCase = 0;
	while (scanf("%d", &n) != EOF&&n)
	{
		int i, j,k;
		iCase++;
		printf("Scenario #%d\n", iCase);
		for (i = 1; i <= n; i++)
		{
			scanf("%d%d",&corr[i].x,&corr[i].y);
		}
		for (i = 1; i <= n; i++)
		{
			for (j = i; j <= n; j++)
			{
				if (i == j) d[i][j] =counter(i,j);
				else d[j][i]=d[i][j] = counter(i, j);
			}
		}
		for (i = 1; i <= n; i++)
		{
			for ( j = 1; j <= n; j++)
			{
				for ( k = 1; k <= n; k++)
				{
					if (d[j][k] > max(d[j][i],d[i][k]))
					{
						d[j][k] = max(d[j][i], d[i][k]);
					}
				}
			}
		}
		printf("Frog Distance = %.3f\n\n", d[1][2]);


	}
}

更新一组测试数据：

2
0 0
3 4

3
17 4
19 4
18 5
8
1 1
4 0
1 2
2 2
3 2
4 2
3 0
5 1
3
9 10
10 10
100 10
6
5 5
100 100
4 4
3 3
2 2
1 1
5
1 2
2 1
3 2
4 1
5 2
3
999 999
1 1
3 3

答案

Scenario #1
Frog Distance = 5.000

Scenario #2
Frog Distance = 1.414

Scenario #3
Frog Distance = 1.414

Scenario #4
Frog Distance = 1.000

Scenario #5
Frog Distance = 134.350

Scenario #6
Frog Distance = 1.414

Scenario #7
Frog Distance = 1408.557

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