POJ-2253 Frogger---最短路变形&&最大边的最小值

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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<sstream>
#define MEM(a, b) memset(a, b, sizeof(a));
using namespace std;
typedef long long ll;
const int maxn = 200 + 10;
const int INF = 0x3f3f3f3f;
int T, n, m, cases, tot;
double Map[maxn][maxn];
struct node
{
    double x, y;
};
node a[maxn];
int main()
{
    while(cin >> n && n)
    {
        MEM(a, 0);
        MEM(Map, 0);
        for(int i = 1; i <= n; i++)
        {
            cin >> a[i].x >> a[i].y;
        }
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= n; j++)
                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));
        }
        for(int k = 1; k <= n; k++)
        {
            for(int i = 1; i <= n; i++)
            {
                for(int j = 1; j <= n; j++)
                {
                    Map[i][j] = min(Map[i][j], max(Map[i][k], Map[k][j]));
                }
            }
        }
        printf("Scenario #%d\n", ++cases);
        printf("Frog Distance = %.3f\n\n", Map[1][2]);
    }
    return 0;
}

dijkstra：47ms

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<sstream>
#define MEM(a, b) memset(a, b, sizeof(a));
using namespace std;
typedef long long ll;
const int maxn = 200 + 10;
const int INF = 0x3f3f3f3f;
int T, n, m, cases, tot;
double Map[maxn][maxn];
struct node
{
    double x, y;
};
node a[maxn];
bool v[maxn];
double d[maxn];
void dijkstra(int u)
{
    memset(v, 0, sizeof(v));
    for(int i = 0; i <= n; i++)d[i] = INF;
    d[u] = 0;
    for(int i = 1; i <= n; i++)
    {
        int x, m = INF;
        for(int i = 1; i <= n; i++)if(!v[i] && d[i] <= m)m = d[x = i];//找最小值
        v[x] = 1;
        for(int i = 1; i <= n; i++)d[i] = min(d[i], max(d[x], Map[x][i]));
    }
}
int main()
{
    while(cin >> n && n)
    {
        MEM(a, 0);
        MEM(Map, 0);
        for(int i = 1; i <= n; i++)
        {
            cin >> a[i].x >> a[i].y;
        }
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= n; j++)
                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));
        }
        dijkstra(1);
        printf("Scenario #%d\n", ++cases);
        printf("Frog Distance = %.3f\n\n", d[2]);
    }
    return 0;
}