POJ

http://poj.org/problem?id=2253

题意：给一些点，完全图，边权是欧几里得距离，最小化从1号点到2号点的路径中的最大边权。

显然的Dijkstra乱搞题，以前还有用Prim改的，虽然Prim和Dijkstra是一个东西。

#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<map>
#include<set>
#include<stack>
#include<string>
#include<queue>
#include<vector>
using namespace std;
typedef long long ll;

double G[205][205];

double dis[205];
bool vis[205];

priority_queue<pair<double, int> > pq;
void dijkstra(int s, int n) {
    while(!pq.empty())
        pq.pop();
    for(int i = 1; i <= n; ++i) {
        dis[i] = 1e18;
        vis[i] = 0;
    }
    dis[s] = 0;
    pq.push({-dis[s], s});
    while(!pq.empty()) {
        int u = pq.top().second;
        pq.pop();
        if(vis[u])
            continue;
        vis[u] = 1;
        for(int v = 1; v <= n; ++v) {
            double w = G[u][v];
            if(!vis[v] && dis[v] > max(dis[u], w)) {
                dis[v] = max(dis[u], w);
                pq.push({-dis[v], v});
            }
        }
    }
}

double x[205], y[205];

int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif // Yinku
    int n, sc = 0;
    while(~scanf("%d", &n)) {
        if(n == 0)
            break;
        for(int i = 1; i <= n; ++i) {
            scanf("%lf%lf", &x[i], &y[i]);
        }
        for(int i = 1; i <= n; ++i) {
            for(int j = 1; j <= n; ++j) {
                G[i][j] = sqrt((x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]));
            }
        }
        dijkstra(1, n);
        printf("Scenario #%d
", ++sc);
        printf("Frog Distance = %.3f

", dis[2]);
    }
}