Highways POJ

题目链接：https://vjudge.net/problem/POJ-1751

思路：

最小生成树板子，只需要多记录每个dis[x]的权值是从哪个点到x这个点的。

#include <stdio.h>
#include <iostream>
#include <queue>
#include <math.h>
#include <algorithm>
using namespace std;

const int N = (int)1e3 + 10;
const int inf = (int)1e9;
double px[N];
double py[N];
double g[N][N];
bool vis[N];
int n,m;
int s;

struct info{
    int from;
    double w;
}dis[N];

struct node{
    int loc;
    double w;
    bool friend operator<(const node & a,const node& b){
        return a.w > b.w;
    }
};
priority_queue<node > que;

inline double dist(int a,int b){
    return sqrt((px[a]-px[b])*(px[a]-px[b])+(py[a]-py[b])*(py[a]-py[b]));
}

void build_map(){

    for(int i = 1; i <= n; i++)
        for(int j = i + 1; j <= n; j++)
            g[i][j] = g[j][i] = dist(i,j);

    for(int i = 1; i <= n; i++) g[i][i] = 0;
}

void prime(){

    for(int i = 1; i <= n; i++){
        vis[i] = 0;
        dis[i].w = inf;
    }

    if(!s) s = 1;

    que.push(node{s,0});
    dis[s].from = 0;
    dis[s].w = 0;

    while(!que.empty()){
        int u = que.top().loc;
        que.pop();
        vis[u] = 1;

        for(int v = 1; v <= n; v++){
            if(!vis[v] && dis[v].w > g[u][v]){
                dis[v].w = g[u][v];
                dis[v].from = u;
                que.push(node{v,dis[v].w});
            }
        }
    }

    for(int i = 1; i <= n; i++){
        if(dis[i].w == 0) continue;
        printf("%d %d
",dis[i].from,i);
    }
}

int main(){


    scanf("%d",&n);

    for(int i = 1; i <= n; i++)
        scanf("%lf%lf",&px[i],&py[i]);

    build_map();

    scanf("%d",&m);
    int u,v;

    for(int i = 1; i <= m ;i++){
        scanf("%d%d",&u,&v);
        s = u;
        g[u][v] = g[v][u] = 0;
    }

    prime();
    return 0;
}