Frogger题解

Frogger

法一：floyd
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<iomanip>
#define int long long int
using namespace std;

const int N = 510;
const int M = 1e5 + 10;
int  n, idx;
double f[N][N], x[M], y[M];

signed main(){
    while(cin >> n && n){
        idx ++;
        memset(f, 0, sizeof f);
        for(int i = 1; i <= n; i++){
            cin >> x[i] >> y[i];
        }
        for(int i = 1;i <= n; i++){
            for(int j = i + 1; j <= n; j++){
                double dis = sqrt((x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]));
                f[i][j] = dis;
                f[j][i] = dis;
            }
        }
        for(int k = 1; k <= n; k++){
            for(int i = 1; i <= n; i++){
                for(int j = 1; j <= n; j++){
                    f[i][j] = min(f[i][j], max(f[i][k], f[k][j]));
                }
            }
        }
        cout <<"Scenario #" << idx << endl;
        cout <<"Frog Distance = " << fixed << setprecision(3) << f[1][2] << endl;
        cout << endl;
    }
    return 0;
}

法二：dijkstra
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<iomanip>
#define int long long int
using namespace std;

const int N = 510;
const int M = 1e5 + 10;
int  n, idx;
double f[N][N], x[M], y[M], dist[N];
bool vis[N];

void dijkstra(){
    for(int i = 1; i <= n; i++){
        vis[i] = false;
        dist[i] = f[1][i];
    }
    vis[1] = true;
    //每次找到未被更新的最小的值对应的点，然后用这个点去更新其他点
    for(int i = 1; i <= n; i++){
        int u = -1;
        double minn = 0x3f3f3f3f;
        for(int j = 1; j <= n; j++){
            if(!vis[j] && dist[j] < minn){
                minn = dist[j];
                u = j;
            }
        }
        vis[u] = true;
        for(int j = 1; j <= n; j++){
            if(!vis[j] && dist[j] > max(dist[u], f[u][j])){
                dist[j] = max(dist[u], f[u][j]);
            }
        }
    }
}

signed main(){
    while(cin >> n && n){
        idx ++;
        memset(f, 0, sizeof f);
        for(int i = 1; i <= n; i++){
            cin >> x[i] >> y[i];
        }
        for(int i = 1;i <= n; i++){
            for(int j = i + 1; j <= n; j++){
                double dis = sqrt((x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]));
                f[i][j] = dis;
                f[j][i] = dis;
            }
        }
        dijkstra();
        cout <<"Scenario #" << idx << endl;
        cout <<"Frog Distance = " << fixed << setprecision(3) << dist[2] << endl;
        cout << endl;
    }
    return 0;
}

原文地址：http://www.cnblogs.com/N-lim/p/16906987.html