POJ 2253 Frogger dijkstra

就是从某点到某点找一条路径，使得这条路径上的最长的长度最短
这个问题，跟最短路的性质类似，所以也可以这样去搞，
if (!vis[j] && max(dist[k], graph[k][j]) < dist[j])

   {

    dist[j] = max(dist[k], graph[k][j]);

    //path[j] = k;

   }
就是dist[j]存储从出发点到该点的路径上的最长边，然后最后搜完就行

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <queue>
using namespace std;
#define ll long long
#define ull unsigned long long
#define INF 0x3f3f3f3f
#define maxn 1005
struct Node
{
int x, y;
};
int Euclid(Node a, Node b)
{
return (a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y);
}
Node stone[maxn];
int graph[maxn][maxn];
//int path[maxn];
bool vis[maxn];
int dist[maxn];
int n, m;
void dijkstra(int s)
{
memset(dist, 0x3f, sizeof(int)*maxn);
memset(vis, false, sizeof(vis));
//memset(path, -1, sizeof(int)*maxn);
dist[s] = 0;
for (int i = 0; i < n; ++i)
{
int k = -1, minv = INF;
for (int j = 0; j < n; ++j)
{
if (!vis[j] && dist[j] < minv)
{
minv = dist[j];
k = j;
}
}
if (k == -1)
break;
vis[k] = true;
//printf("k %d\n", k);
for (int j = 0; j < n; ++j)
{
if (!vis[j] && max(dist[k], graph[k][j]) < dist[j])
{
dist[j] = max(dist[k], graph[k][j]);
//path[j] = k;
}
}
}
}
int main()
{
//freopen("input.txt", "r", stdin);
int kase = 0;
while (scanf("%d", &n) != EOF)
{
if (n == 0)
break;
memset(graph, 0x3f, sizeof(int)*maxn*maxn);
for (int i = 0; i < n; ++i)
{
scanf("%d%d", &stone[i].x, &stone[i].y);
}
for (int i = 0; i < n; ++i)
{
for (int j = i + 1; j < n; ++j)
{
graph[i][j] = Euclid(stone[i], stone[j]);
graph[j][i] = graph[i][j];
}
}
/*for (int i = 0; i < n; ++i)
{
for (int j = 0; j < n; ++j)
printf("%d ", graph[i][j]);
printf("\n");
}*/
dijkstra(0);
int ans = dist[1];
printf("Scenario #%d\nFrog Distance = %.3f\n\n", ++kase, sqrt(ans*1.0));
}
//system("pause");
//while (1);
return 0;
}